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Class XII Chapter 7 – Integrals Maths
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Exercise 7.1
Question 1:
sin 2x
Answer
The anti derivative of sin 2x is a function of x whose derivative is sin 2x.It is known that,
Therefore, the anti derivative of
Question 2:
Cos 3x
Answer
The anti derivative of cos 3x is a function of x whose derivative is cos 3x.
It is known that,
Therefore, the anti derivative of .
Question 3:
e2x
Answer
The anti derivative of e2x is the function of x whose derivative is e2x.
It is known that,
Class XII Chapter 7 – Integrals Maths
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Therefore, the anti derivative of .
Question 4:
Answer
The anti derivative of is the function of x whose derivative is .
It is known that,
Therefore, the anti derivative of .
Question 5:
Answer
The anti derivative of is the function of x whose derivative is
.
It is known that,
Class XII Chapter 7 – Integrals Maths
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Therefore, the anti derivative of is .
Question 6:
Answer
Question 7:
Answer
Question 8:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 9:
Answer
Question 10:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 11:
Answer
Question 12:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 13:
Answer
On dividing, we obtain
Question 14:
Answer
Question 15:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 16:
Answer
Question 17:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 18:
Answer
Question 19:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 20:
Answer
Question 21:
The anti derivative of equals
(A) (B)
(C) (D)
Answer
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is C.
Question 22:
If such that f(2) = 0, then f(x) is
(A) (B)
(C) (D)
Answer
It is given that,
∴Anti derivative of
Class XII Chapter 7 – Integrals Maths
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∴
Also,
Hence, the correct Answer is A.
Class XII Chapter 7 – Integrals Maths
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Exercise 7.2
Question 1:
Answer
Let = t
∴2x dx = dt
Question 2:
Answer
Let log |x| = t
∴
Class XII Chapter 7 – Integrals Maths
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Question 3:
Answer
Let 1 + log x = t
∴
Question 4:
sin x ⋅ sin (cos x)
Answer
Class XII Chapter 7 – Integrals Maths
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sin x ⋅ sin (cos x)
Let cos x = t
∴ −sin x dx = dt
Question 5:
Answer
Let
∴ 2adx = dt
Class XII Chapter 7 – Integrals Maths
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Question 6:
Answer
Let ax + b = t
⇒ adx = dt
Question 7:
Answer
Let
Class XII Chapter 7 – Integrals Maths
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∴ dx = dt
Question 8:
Answer
Let 1 + 2x2 = t
∴ 4xdx = dt
Class XII Chapter 7 – Integrals Maths
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Question 9:
Answer
Let
∴ (2x + 1)dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 10:
Answer
Let
∴
Question 11:
Answer
Let
∴ dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 12:
Answer
Let
∴
Class XII Chapter 7 – Integrals Maths
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Question 13:
Answer
Let
∴ 9x2 dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 14:
Answer
Let log x = t
∴
Question 15:
Answer
Class XII Chapter 7 – Integrals Maths
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Let
∴ −8x dx = dt
Question 16:
Answer
Let
∴ 2dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 17:
Answer
Let
∴ 2xdx = dt
Question 18:
Answer
Let
∴
Class XII Chapter 7 – Integrals Maths
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Question 19:
Answer
Dividing numerator and denominator by ex, we obtain
Let
∴
Class XII Chapter 7 – Integrals Maths
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Question 20:
Answer
Let
∴
Question 21:
Answer
Let 2x − 3 = t
∴ 2dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 22:
Answer
Let 7 − 4x = t
∴ −4dx = dt
Question 23:
Answer
Let
Class XII Chapter 7 – Integrals Maths
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∴
Question 24:
Answer
Let
∴
Class XII Chapter 7 – Integrals Maths
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Question 25:
Answer
Let
∴
Question 26:
Answer
Let
∴
Class XII Chapter 7 – Integrals Maths
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Question 27:
Answer
Let sin 2x = t
∴
Question 28:
Answer
Let
Class XII Chapter 7 – Integrals Maths
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∴ cos x dx = dt
Question 29:
cot x log sin x
Answer
Let log sin x = t
Question 30:
Answer
Let 1 + cos x = t
Class XII Chapter 7 – Integrals Maths
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∴ −sin x dx = dt
Question 31:
Answer
Let 1 + cos x = t
∴ −sin x dx = dt
Question 32:
Answer
Class XII Chapter 7 – Integrals Maths
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Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt
Question 33:
Answer
Class XII Chapter 7 – Integrals Maths
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Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt
Question 34:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 35:
Answer
Let 1 + log x = t
∴
Class XII Chapter 7 – Integrals Maths
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Question 36:
Answer
Let
∴
Question 37:
Answer
Let x4 = t
∴ 4x3 dx = dt
Class XII Chapter 7 – Integrals Maths
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Let
∴
From (1), we obtain
Question 38:
equals
Answer
Let
∴
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is D.
Question 39:
equals
A.
B.
C.
D.
Answer
Hence, the correct Answer is B.
Class XII Chapter 7 – Integrals Maths
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Exercise 7.3
Question 1:
Answer
Question 2:
Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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Question 3:
cos 2x cos 4x cos 6x
Answer
It is known that,
Question 4:
sin3 (2x + 1)
Answer
Let
Class XII Chapter 7 – Integrals Maths
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Question 5:
sin3 x cos3 x
Answer
Question 6:
sin x sin 2x sin 3x
Answer
Class XII Chapter 7 – Integrals Maths
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It is known that,
Question 7:
sin 4x sin 8x
Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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Question 8:
Answer
Question 9:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 10:
sin4 x
Answer
Class XII Chapter 7 – Integrals Maths
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Question 11:
cos4 2x
Answer
Class XII Chapter 7 – Integrals Maths
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Question 12:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 13:
Answer
Question 14:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 15:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 16:
tan4x
Answer
Class XII Chapter 7 – Integrals Maths
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From equation (1), we obtain
Question 17:
Answer
Question 18:
Answer
Question 19:
Class XII Chapter 7 – Integrals Maths
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Answer
Question 20:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 21:
sin−1 (cos x)
Answer
Class XII Chapter 7 – Integrals Maths
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It is known that,
Substituting in equation (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 22:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 23:
is equal to
A. tan x + cot x + C
B. tan x + cosec x + C
C. − tan x + cot x + C
D. tan x + sec x + C
Answer
Hence, the correct Answer is A.
Question 24:
equals
A. − cot (exx) + C
B. tan (xex) + C
C. tan (ex) + C
D. cot (ex) + C
Answer
Let exx = t
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is B.
Class XII Chapter 7 – Integrals Maths
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Exercise 7.4
Question 1:
Answer
Let x3 = t
∴ 3x2 dx = dt
Question 2:
Answer
Let 2x = t
∴ 2dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 3:
Answer
Let 2 − x = t
⇒ −dx = dt
Question 4:
Answer
Let 5x = t
Class XII Chapter 7 – Integrals Maths
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∴ 5dx = dt
Question 5:
Answer
Question 6:
Answer
Let x3 = t
Class XII Chapter 7 – Integrals Maths
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∴ 3x2 dx = dt
Question 7:
Answer
From (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 8:
Answer
Let x3 = t
⇒ 3x2 dx = dt
Question 9:
Answer
Let tan x = t
∴ sec2x dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 10:
Answer
Question 11:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 12:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 13:
Answer
Question 14:
Class XII Chapter 7 – Integrals Maths
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Answer
Question 15:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 16:
Answer
Equating the coefficients of x and constant term on both sides, we obtain
Class XII Chapter 7 – Integrals Maths
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4A = 4 ⇒ A = 1
A + B = 1 ⇒ B = 0
Let 2x2 + x − 3 = t
∴ (4x + 1) dx = dt
Question 17:
Answer
Equating the coefficients of x and constant term on both sides, we obtain
From (1), we obtain
Class XII Chapter 7 – Integrals Maths
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From equation (2), we obtain
Question 18:
Answer
Equating the coefficient of x and constant term on both sides, we obtain
Class XII Chapter 7 – Integrals Maths
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Class XII Chapter 7 – Integrals Maths
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Substituting equations (2) and (3) in equation (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 19:
Answer
Equating the coefficients of x and constant term, we obtain
2A = 6 ⇒ A = 3
−9A + B = 7 ⇒ B = 34
∴ 6x + 7 = 3 (2x − 9) + 34
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Substituting equations (2) and (3) in (1), we obtain
Question 20:
Answer
Equating the coefficients of x and constant term on both sides, we obtain
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Using equations (2) and (3) in (1), we obtain
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Question 21:
Answer
Let x2 + 2x +3 = t
⇒ (2x + 2) dx =dt
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Using equations (2) and (3) in (1), we obtain
Question 22:
Answer
Equating the coefficients of x and constant term on both sides, we obtain
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Substituting (2) and (3) in (1), we obtain
Question 23:
Answer
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Equating the coefficients of x and constant term, we obtain
Using equations (2) and (3) in (1), we obtain
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Question 24:
equals
A. x tan−1 (x + 1) + C
B. tan− 1 (x + 1) + C
C. (x + 1) tan−1 x + C
D. tan−1 x + C
Answer
Hence, the correct Answer is B.
Question 25:
equals
A.
B.
C.
Class XII Chapter 7 – Integrals Maths
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D.
Answer
Hence, the correct Answer is B.
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Exercise 7.5
Question 1:
Answer
Let
Equating the coefficients of x and constant term, we obtain
A + B = 1
2A + B = 0
On solving, we obtain
A = −1 and B = 2
Question 2:
Answer
Let
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Equating the coefficients of x and constant term, we obtain
A + B = 0
−3A + 3B = 1
On solving, we obtain
Question 3:
Answer
Let
Substituting x = 1, 2, and 3 respectively in equation (1), we obtain
A = 1, B = −5, and C = 4
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Question 4:
Answer
Let
Substituting x = 1, 2, and 3 respectively in equation (1), we obtain
Question 5:
Answer
Let
Substituting x = −1 and −2 in equation (1), we obtain
A = −2 and B = 4
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Question 6:
Answer
It can be seen that the given integrand is not a proper fraction.
Therefore, on dividing (1 − x2) by x(1 − 2x), we obtain
Let
Substituting x = 0 and in equation (1), we obtain
A = 2 and B = 3
Substituting in equation (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 7:
Answer
Let
Equating the coefficients of x2, x, and constant term, we obtain
A + C = 0
−A + B = 1
−B + C = 0
On solving these equations, we obtain
From equation (1), we obtain
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Question 8:
Answer
Let
Substituting x = 1, we obtain
Equating the coefficients of x2 and constant term, we obtain
A + C = 0
−2A + 2B + C = 0
On solving, we obtain
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Question 9:
Answer
Let
Substituting x = 1 in equation (1), we obtain
B = 4
Equating the coefficients of x2 and x, we obtain
A + C = 0
B − 2C = 3
On solving, we obtain
Class XII Chapter 7 – Integrals Maths
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Question 10:
Answer
Let
Equating the coefficients of x2 and x, we obtain
Class XII Chapter 7 – Integrals Maths
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Question 11:
Answer
Let
Substituting x = −1, −2, and 2 respectively in equation (1), we obtain
Question 12:
Class XII Chapter 7 – Integrals Maths
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Answer
It can be seen that the given integrand is not a proper fraction.
Therefore, on dividing (x3 + x + 1) by x2 − 1, we obtain
Let
Substituting x = 1 and −1 in equation (1), we obtain
Question 13:
Answer
Equating the coefficient of x2, x, and constant term, we obtain
A − B = 0
B − C = 0
A + C = 2
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On solving these equations, we obtain
A = 1, B = 1, and C = 1
Question 14:
Answer
Equating the coefficient of x and constant term, we obtain
A = 3
2A + B = −1 ⇒ B = −7
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Question 15:
Answer
Equating the coefficient of x3, x2, x, and constant term, we obtain
On solving these equations, we obtain
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Question 16:
[Hint: multiply numerator and denominator by xn − 1 and put xn = t]
Answer
Multiplying numerator and denominator by xn − 1, we obtain
Substituting t = 0, −1 in equation (1), we obtain
A = 1 and B = −1
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Question 17:
[Hint: Put sin x = t]
Answer
Substituting t = 2 and then t = 1 in equation (1), we obtain
A = 1 and B = −1
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Question 18:
Answer
Equating the coefficients of x3, x2, x, and constant term, we obtain
A + C = 0
B + D = 4
4A + 3C = 0
4B + 3D = 10
On solving these equations, we obtain
A = 0, B = −2, C = 0, and D = 6
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Question 19:
Answer
Let x2 = t ⇒ 2x dx = dt
Substituting t = −3 and t = −1 in equation (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 20:
Answer
Multiplying numerator and denominator by x3, we obtain
Let x4 = t ⇒ 4x3dx = dt
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Substituting t = 0 and 1 in (1), we obtain
A = −1 and B = 1
Question 21:
[Hint: Put ex = t]
Answer
Let ex = t ⇒ ex dx = dt
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Substituting t = 1 and t = 0 in equation (1), we obtain
A = −1 and B = 1
Question 22:
A.
B.
C.
D.
Answer
Substituting x = 1 and 2 in (1), we obtain
A = −1 and B = 2
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is B.
Question 23:
A.
B.
C.
D.
Answer
Equating the coefficients of x2, x, and constant term, we obtain
A + B = 0
C = 0
A = 1
On solving these equations, we obtain
A = 1, B = −1, and C = 0
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is A.
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Exercise 7.6
Question 1:
x sin x
Answer
Let I =
Taking x as first function and sin x as second function and integrating by parts, we
obtain
Question 2:
Answer
Let I =
Taking x as first function and sin 3x as second function and integrating by parts, we
obtain
Question 3:
Class XII Chapter 7 – Integrals Maths
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Answer
Let
Taking x2 as first function and ex as second function and integrating by parts, we obtain
Again integrating by parts, we obtain
Question 4:
x logx
Answer
Let
Taking log x as first function and x as second function and integrating by parts, we
obtain
Class XII Chapter 7 – Integrals Maths
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Question 5:
x log 2x
Answer
Let
Taking log 2x as first function and x as second function and integrating by parts, we
obtain
Question 6:
x2 log x
Answer
Let
Taking log x as first function and x2 as second function and integrating by parts, we
obtain
Class XII Chapter 7 – Integrals Maths
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Question 7:
Answer
Let
Taking as first function and x as second function and integrating by parts, we
obtain
Question 8:
Answer
Let
Class XII Chapter 7 – Integrals Maths
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Taking as first function and x as second function and integrating by parts, we
obtain
Question 9:
Answer
Let
Taking cos−1 x as first function and x as second function and integrating by parts, we
obtain
Class XII Chapter 7 – Integrals Maths
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Class XII Chapter 7 – Integrals Maths
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Question 10:
Answer
Let
Taking as first function and 1 as second function and integrating by parts, we
obtain
Question 11:
Answer
Let
Class XII Chapter 7 – Integrals Maths
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Taking as first function and as second function and integrating by parts,
we obtain
Question 12:
Answer
Let
Taking x as first function and sec2x as second function and integrating by parts, we
obtain
Question 13:
Answer
Class XII Chapter 7 – Integrals Maths
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Let
Taking as first function and 1 as second function and integrating by parts, we
obtain
Question 14:
Answer
Taking as first function and 1 as second function and integrating by parts, we
obtain
Again integrating by parts, we obtain
Class XII Chapter 7 – Integrals Maths
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Question 15:
Answer
Let
Let I = I1 + I2 … (1)
Where, and
Taking log x as first function and x2 as second function and integrating by parts, we
obtain
Taking log x as first function and 1 as second function and integrating by parts, we
obtain
Class XII Chapter 7 – Integrals Maths
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Using equations (2) and (3) in (1), we obtain
Question 16:
Answer
Let
Let
⇒
⇒
It is known that,
Question 17:
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Answer
Let
Let ⇒
It is known that,
Question 18:
Answer
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Let ⇒
It is known that,
From equation (1), we obtain
Question 19:
Class XII Chapter 7 – Integrals Maths
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Answer
Also, let ⇒
It is known that,
Question 20:
Answer
Let ⇒
It is known that,
Question 21:
Answer
Class XII Chapter 7 – Integrals Maths
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Let
Integrating by parts, we obtain
Again integrating by parts, we obtain
Question 22:
Answer
Let ⇒
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= 2θ
⇒
Integrating by parts, we obtain
Question 23:
equals
Answer
Let
Also, let ⇒
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is A.
Question 24:
equals
Answer
Let
Also, let ⇒
It is known that,
Hence, the correct Answer is B.
Class XII Chapter 7 – Integrals Maths
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Exercise 7.7
Question 1:
Answer
Question 2:
Answer
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Question 3:
Answer
Question 4:
Answer
Question 5:
Class XII Chapter 7 – Integrals Maths
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Answer
Question 6:
Answer
Question 7:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 8:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 9:
Answer
Question 10:
is equal to
A.
B.
C.
Class XII Chapter 7 – Integrals Maths
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D.
Answer
Hence, the correct Answer is A.
Question 11:
is equal to
A.
B.
C.
D.
Answer
Hence, the correct Answer is D.
Class XII Chapter 7 – Integrals Maths
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Exercise 7.8
Question 1:
Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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Question 2:
Answer
It is known that,
Question 3:
Class XII Chapter 7 – Integrals Maths
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Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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Question 4:
Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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From equations (2) and (3), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 5:
Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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Question 6:
Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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Class XII Chapter 7 – Integrals Maths
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Exercise 7.9
Question 1:
Answer
By second fundamental theorem of calculus, we obtain
Question 2:
Answer
By second fundamental theorem of calculus, we obtain
Question 3:
Class XII Chapter 7 – Integrals Maths
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Answer
By second fundamental theorem of calculus, we obtain
Question 4:
Answer
By second fundamental theorem of calculus, we obtain
Class XII Chapter 7 – Integrals Maths
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Question 5:
Answer
By second fundamental theorem of calculus, we obtain
Question 6:
Answer
Class XII Chapter 7 – Integrals Maths
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By second fundamental theorem of calculus, we obtain
Question 7:
Answer
By second fundamental theorem of calculus, we obtain
Question 8:
Answer
Class XII Chapter 7 – Integrals Maths
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By second fundamental theorem of calculus, we obtain
Question 9:
Answer
By second fundamental theorem of calculus, we obtain
Question 10:
Class XII Chapter 7 – Integrals Maths
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Answer
By second fundamental theorem of calculus, we obtain
Question 11:
Answer
By second fundamental theorem of calculus, we obtain
Question 12:
Class XII Chapter 7 – Integrals Maths
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Answer
By second fundamental theorem of calculus, we obtain
Question 13:
Answer
By second fundamental theorem of calculus, we obtain
Class XII Chapter 7 – Integrals Maths
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Question 14:
Answer
By second fundamental theorem of calculus, we obtain
Question 15:
Answer
Class XII Chapter 7 – Integrals Maths
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By second fundamental theorem of calculus, we obtain
Question 16:
Answer
Let
Class XII Chapter 7 – Integrals Maths
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Equating the coefficients of x and constant term, we obtain
A = 10 and B = −25
Substituting the value of I1 in (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 17:
Answer
By second fundamental theorem of calculus, we obtain
Question 18:
Answer
Class XII Chapter 7 – Integrals Maths
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By second fundamental theorem of calculus, we obtain
Question 19:
Answer
By second fundamental theorem of calculus, we obtain
Class XII Chapter 7 – Integrals Maths
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Question 20:
Answer
By second fundamental theorem of calculus, we obtain
Class XII Chapter 7 – Integrals Maths
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Question 21:
equals
A.
B.
C.
D.
Answer
By second fundamental theorem of calculus, we obtain
Hence, the correct Answer is D.
Class XII Chapter 7 – Integrals Maths
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Question 22:
equals
A.
B.
C.
D.
Answer
By second fundamental theorem of calculus, we obtain
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is C.
Class XII Chapter 7 – Integrals Maths
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Exercise 7.10
Question 1:
Answer
When x = 0, t = 1 and when x = 1, t = 2
Question 2:
Answer
Also, let
Class XII Chapter 7 – Integrals Maths
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Question 3:
Answer
Also, let x = tanθ ⇒ dx = sec2θ dθ
When x = 0, θ = 0 and when x = 1,
Class XII Chapter 7 – Integrals Maths
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Takingθas first function and sec2θ as second function and integrating by parts, we obtain
Question 4:
Answer
Let x + 2 = t2 ⇒ dx = 2tdt
When x = 0, and when x = 2, t = 2
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Question 5:
Answer
Let cos x = t ⇒ −sinx dx = dt
When x = 0, t = 1 and when
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Question 6:
Answer
Let ⇒ dx = dt
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Question 7:
Answer
Let x + 1 = t ⇒ dx = dt
When x = −1, t = 0 and when x = 1, t = 2
Question 8:
Answer
Let 2x = t ⇒ 2dx = dt
When x = 1, t = 2 and when x = 2, t = 4
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Question 9:
The value of the integral is
A. 6
B. 0
C. 3
D. 4
Answer
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Let cotθ = t ⇒ −cosec2θ dθ= dt
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is A.
Question 10:
If
A. cos x + x sin x
B. x sin x
C. x cos x
D. sin x + x cos x
Answer
Integrating by parts, we obtain
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is B.
Class XII Chapter 7 – Integrals Maths
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Exercise 7.11
Question 1:
Answer
Adding (1) and (2), we obtain
Question 2:
Answer
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Adding (1) and (2), we obtain
Question 3:
Answer
Class XII Chapter 7 – Integrals Maths
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Adding (1) and (2), we obtain
Question 4:
Answer
Class XII Chapter 7 – Integrals Maths
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Adding (1) and (2), we obtain
Question 5:
Answer
It can be seen that (x + 2) ≤ 0 on [−5, −2] and (x + 2) ≥ 0 on [−2, 5].
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Question 6:
Answer
It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8].
Question 7:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 8:
Answer
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Question 9:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 10:
Answer
Class XII Chapter 7 – Integrals Maths
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Adding (1) and (2), we obtain
Question 11:
Answer
As sin2 (−x) = (sin (−x))2 = (−sin x)2 = sin2x, therefore, sin2x is an even function.
Class XII Chapter 7 – Integrals Maths
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It is known that if f(x) is an even function, then
Question 12:
Answer
Adding (1) and (2), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 13:
Answer
As sin7 (−x) = (sin (−x))7 = (−sin x)7 = −sin7x, therefore, sin2x is an odd function.
It is known that, if f(x) is an odd function, then
Question 14:
Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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Question 15:
Answer
Adding (1) and (2), we obtain
Question 16:
Answer
Class XII Chapter 7 – Integrals Maths
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Adding (1) and (2), we obtain
sin (π − x) = sin x
Adding (4) and (5), we obtain
Let 2x = t ⇒ 2dx = dt
When x = 0, t = 0 and when
Class XII Chapter 7 – Integrals Maths
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Question 17:
Answer
It is known that,
Adding (1) and (2), we obtain
Question 18:
Answer
It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4
Class XII Chapter 7 – Integrals Maths
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Question 19:
Show that if f and g are defined as and
Answer
Adding (1) and (2), we obtain
Question 20:
Class XII Chapter 7 – Integrals Maths
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The value of is
A. 0
B. 2
C. π
D. 1
Answer
It is known that if f(x) is an even function, then and
if f(x) is an odd function, then
Hence, the correct Answer is C.
Question 21:
The value of is
A. 2
B.
C. 0
D.
Class XII Chapter 7 – Integrals Maths
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Answer
Adding (1) and (2), we obtain
Hence, the correct Answer is C.
Class XII Chapter 7 – Integrals Maths
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Miscellaneous Solutions
Question 1:
Answer
Equating the coefficients of x2, x, and constant term, we obtain
−A + B − C = 0
B + C = 0
A = 1
On solving these equations, we obtain
From equation (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 2:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 3:
[Hint: Put ]
Answer
Class XII Chapter 7 – Integrals Maths
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Question 4:
Answer
Class XII Chapter 7 – Integrals Maths
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Class XII Chapter 7 – Integrals Maths
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Question 5:
Answer
On dividing, we obtain
Question 6:
Class XII Chapter 7 – Integrals Maths
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Answer
Equating the coefficients of x2, x, and constant term, we obtain
A + B = 0
B + C = 5
9A + C = 0
On solving these equations, we obtain
From equation (1), we obtain
Question 7:
Answer
Class XII Chapter 7 – Integrals Maths
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Let x − a = t ⇒ dx = dt
Question 8:
Answer
Question 9:
Answer
Class XII Chapter 7 – Integrals Maths
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Let sin x = t ⇒ cos x dx = dt
Question 10:
Answer
Question 11:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 12:
Answer
Let x4 = t ⇒ 4x3 dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 13:
Answer
Let ex = t ⇒ ex dx = dt
Question 14:
Answer
Class XII Chapter 7 – Integrals Maths
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Equating the coefficients of x3, x2, x, and constant term, we obtain
A + C = 0
B + D = 0
4A + C = 0
4B + D = 1
On solving these equations, we obtain
From equation (1), we obtain
Question 15:
Answer
= cos3 x × sin x
Let cos x = t ⇒ −sin x dx = dt
Class XII Chapter 7 – Integrals Maths
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Question 16:
Answer
Question 17:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 18:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 19:
Answer
Class XII Chapter 7 – Integrals Maths
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From equation (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 20:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 21:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 22:
Answer
Equating the coefficients of x2, x,and constant term, we obtain
A + C = 1
3A + B + 2C = 1
2A + 2B + C = 1
On solving these equations, we obtain
A = −2, B = 1, and C = 3
From equation (1), we obtain
Class XII Chapter 7 – Integrals Maths
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Question 23:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 24:
Answer
Integrating by parts, we obtain
Class XII Chapter 7 – Integrals Maths
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Question 25:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 26:
Answer
Class XII Chapter 7 – Integrals Maths
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When x = 0, t = 0 and
Question 27:
Answer
Class XII Chapter 7 – Integrals Maths
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When and when
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Question 28:
Answer
When and when
As , therefore, is an even function.
It is known that if f(x) is an even function, then
Class XII Chapter 7 – Integrals Maths
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Question 29:
Answer
Class XII Chapter 7 – Integrals Maths
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Question 30:
Answer
Question 31:
Answer
Class XII Chapter 7 – Integrals Maths
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From equation (1), we obtain
Question 32:
Answer
Class XII Chapter 7 – Integrals Maths
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Adding (1) and (2), we obtain
Question 33:
Class XII Chapter 7 – Integrals Maths
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Answer
Class XII Chapter 7 – Integrals Maths
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From equations (1), (2), (3), and (4), we obtain
Question 34:
Answer
Equating the coefficients of x2, x, and constant term, we obtain
A + C = 0
A + B = 0
B = 1
On solving these equations, we obtain
A = −1, C = 1, and B = 1
Class XII Chapter 7 – Integrals Maths
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Hence, the given result is proved.
Question 35:
Answer
Integrating by parts, we obtain
Class XII Chapter 7 – Integrals Maths
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Hence, the given result is proved.
Question 36:
Answer
Therefore, f (x) is an odd function.
It is known that if f(x) is an odd function, then
Hence, the given result is proved.
Question 37:
Answer
Class XII Chapter 7 – Integrals Maths
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Hence, the given result is proved.
Question 38:
Answer
Hence, the given result is proved.
Question 39:
Answer
Integrating by parts, we obtain
Class XII Chapter 7 – Integrals Maths
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Let 1 − x2 = t ⇒ −2x dx = dt
Hence, the given result is proved.
Question 40:
Evaluate as a limit of a sum.
Answer
It is known that,
Class XII Chapter 7 – Integrals Maths
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Question 41:
is equal to
A.
Class XII Chapter 7 – Integrals Maths
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B.
C.
D.
Answer
Hence, the correct Answer is A.
Question 42:
is equal to
A.
B.
C.
D.
Answer
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is B.
Question 43:
If then is equal to
A.
B.
C.
D.
Answer
Class XII Chapter 7 – Integrals Maths
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Hence, the correct Answer is D.
Question 44:
The value of is
A. 1
B. 0
C. − 1
D.
Answer
Class XII Chapter 7 – Integrals Maths
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Adding (1) and (2), we obtain
Hence, the correct Answer is B.
27
(ii) The differential of y, denoted by dy , is defined by dxxfdy 0' or
xdx
dydy
.
PRACTICE PROBLEMS
LEVEL -I
Find the slope of the tangent to the curve xxy 43 2 at x = 4
Find the slope of the tangent to the curve 𝑦 = 𝑥2 − 3𝑥 + 2 at the point whose x-
coordinate is 3
Find the slope of the normal to the curve 4
,,sin,cos 33 atayax
LEVEL –II
Find the point at which the tangent to the curve 𝑦 = √4𝑥 − 3 − 1 has its slope 2
3.
Find the equation of the tangent line to the curve 722 xxy which is (i)
parallel to the line 092 yx (ii) Perpendicular to the line 13155 xy
Find the point on the curve 03222 xyx at which the tangents are the
parallel to the x-axis.
LEVEL-III
Find the equation of the normal to the curve 623 xxy which are parallel to the
line x + 14y + 4 = 0.
Prove that the curves 2yx and kxy cut at right angleif 18 3 k
Find the equation of the tangent to the curve 23 xy which is parallel to the line
0524 yx
INTEGRATION
INTRODUCTION
IF f(x) is derivative of function g(x), then g(x) is known as antiderivative or integral of
f(x)
i.e., 𝒅
𝒅𝒙 (g(x)) = f(x) ⇔ ∫ 𝒇(𝒙)𝒅𝒙 = 𝒈(𝒙)
STANDARD SET OF FORMULAS
* Where c is an arbitrary constant.
1. ∫𝒙𝒏 𝒅𝒙 = 𝒙𝒏+𝟏
𝒏+𝟏 + 𝒄 (n -1)
2. ∫𝒅𝒙 = x + c
3. ∫𝟏
𝒙 dx = log |x| + c
4. ∫ 𝒄𝒐𝒔 𝒙 𝒅𝒙 = 𝒔𝒊𝒏 𝒙 + 𝒄
28
5. ∫ 𝒔𝒊𝒏 𝒙 𝒅𝒙 = − 𝒄𝒐𝒔 𝒙 + 𝒄
6. ∫ 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 = 𝒕𝒂𝒏 𝒙 + 𝒄
7. ∫ 𝒄𝒐𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 = −𝒄𝒐𝒕 𝒙 + 𝒄
8. ∫ 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 𝒅𝒙 = 𝒔𝒆𝒄 𝒙 + 𝒄
9. ∫ 𝒄𝒐𝒔𝒆𝒄 𝒙 𝒄𝒐𝒕 𝒙 𝒅𝒙 = −𝒄𝒐𝒔𝒆𝒄 𝒙 + 𝒄
10. ∫𝒆𝒙 𝒅𝒙 = 𝒆𝒙 + 𝒄
11. ∫ 𝒕𝒂𝒏 𝒙 𝒅𝒙 = 𝒍𝒐𝒈 |𝒔𝒆𝒄 𝒙 | + 𝒄
12. ∫ 𝒄𝒐𝒕 𝒙 𝒅𝒙 = 𝒍𝒐𝒈 |𝒔𝒊𝒏 𝒙 | + 𝒄
13. ∫ 𝒔𝒆𝒄 𝒙 𝒅𝒙 = 𝒍𝒐𝒈 |𝒔𝒆𝒄 𝒙 + 𝒕𝒂𝒏 𝒙 | + 𝒄
14. ∫ 𝒄𝒐𝒔𝒆𝒄 𝒙 𝒅𝒙 = 𝒍𝒐𝒈 |𝒄𝒐𝒔𝒆𝒄 𝒙 − 𝒄𝒐𝒕 𝒙 | + 𝒄
15. ∫𝟏
√𝟏− 𝒙𝟐 dx = sin -1 x + c
16. ∫𝟏
𝟏+𝒙𝟐 𝒅𝒙 = 𝒕𝒂𝒏−𝟏𝒙 + 𝒄
17. ∫𝟏
𝒙√𝒙𝟐− 𝟏 𝒅𝒙 = 𝐬𝐞𝐜−𝟏 𝒙 + 𝒄
18. ∫𝒂𝒙 𝒅𝒙 = 𝒂𝒙
𝒍𝒐𝒈 𝒂+ 𝒄
19. ∫𝟏
√𝒙 𝒅𝒙 = 𝟐√𝒙 + 𝒄
INTEGRALS OF LINEAR FUNCTIONS
1. ∫(𝒂𝒙 + 𝒃)𝒏 𝒅𝒙 = (𝒂𝒙+𝒃)𝒏+𝟏
(𝒏+𝟏)𝒂+ 𝒄
2. ∫𝟏
𝒂𝒙+𝒃 𝒅𝒙 =
𝒍𝒐𝒈 (𝒂𝒙+𝒃)
𝒂+ 𝒄
3. ∫ 𝒔𝒊𝒏 (𝒂𝒙 + 𝒃 )𝒅𝒙 = −𝒄𝒐𝒔 (𝒂𝒙+𝒃)
𝒂+ 𝒄
In the same way if ax +b comes in the place of x, in the standard set of formulas, then
divide the integral by a
SPECIAL INTEGRALS
1. ∫𝟏
𝒙𝟐−𝒂𝟐 𝒅𝒙 = 𝟏
𝟐𝒂 𝒍𝒐𝒈 |
𝒙−𝒂
𝒙+𝒂| + c
2. ∫𝟏
𝒂𝟐−𝒙𝟐 𝒅𝒙 = 𝟏
𝟐𝒂 𝒍𝒐𝒈 |
𝒂+𝒙
𝒂−𝒙| + 𝒄
3. ca
x
adx
ax
1
22tan
11
4. ∫𝟏
√𝒙𝟐−𝒂𝟐 𝒅𝒙 = 𝒍𝒐𝒈| 𝒙 + √𝒙𝟐 − 𝒂𝟐 | + 𝒄
5. ∫𝟏
√𝒙𝟐+𝒂𝟐 𝒅𝒙 = 𝒍𝒐𝒈 |𝒙 + √𝒙𝟐 + 𝒂𝟐| + c
6. ∫𝟏
√𝒂𝟐−𝒙𝟐 𝒅𝒙 = 𝐬𝐢𝐧−𝟏 (
𝒙
𝒂) + 𝒄
7. ∫√𝒙𝟐 + 𝒂𝟐 𝒅𝒙 =𝒙
𝟐√𝒙𝟐 + 𝒂𝟐 +
𝒂𝟐
𝟐 𝒍𝒐𝒈 | 𝒙 + √𝒙𝟐 + 𝒂𝟐| + 𝒄
8. ∫√𝒙𝟐 − 𝒂𝟐 dx = 𝒙
𝟐√𝒙𝟐 − 𝒂𝟐 −
𝒂𝟐
𝟐 𝒍𝒐𝒈 | 𝒙 + √𝒙𝟐 − 𝒂𝟐| + 𝒄
9. ∫√𝒂𝟐 − 𝒙𝟐 dx = 𝒙
𝟐√𝒂𝟐 − 𝒙𝟐 +
𝒂𝟐
𝟐𝐬𝐢𝐧−𝟏 (
𝒙
𝒂) + 𝒄
INTEGRATION BY PARTS
1. ∫𝒖. 𝒗 𝒅𝒙 = 𝒖. ∫ 𝒗 𝒅𝒙 − ∫( 𝒅𝒖
𝒅𝒙∫𝒗 𝒅𝒙) 𝒅𝒙
29
OR
The integral of product of two functions = (first function) x integral of the second
function – integral of [(differential coefficient of the first function ) × (integral of the
second function)]
We can choose first and second function according to I L A T E where I → inverse
trigonometric function → 𝒍𝒐𝒈𝒂𝒓𝒊𝒕𝒉𝒎𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏, 𝐀 → 𝒂𝒍𝒈𝒆𝒃𝒓𝒂𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝐓 →
𝒕𝒓𝒊𝒈𝒐𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝐄 → 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒕𝒊𝒂𝒍 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
2. cxfedxxfxfe xx )()()( 1 .
Working Rule for different types of integrals
1. Integration of trigonometric function
Working Rule
(a) Express the given integrand as the algebraic sum of the functions of the following
forms
(i) Sin k𝜶 , (ii) cos k𝜶 ,(iii) tan k𝜶 , (iv) cot k𝜶 , (v) sec k𝜶 ,(vi) cosec k𝜶 ,(vii) sec2
k𝜶 ,
(viii) cosec2 k𝜶 , (ix) sec k𝜶 tan k𝜶 (x) cosec k𝜶 cot k𝜶
For this use the following formulae whichever applicable
(i) 𝒔𝒊𝒏𝟐 𝒙 = 𝟏−𝒄𝒐𝒔 𝟐𝒙
𝟐 (ii) 𝒄𝒐𝒔𝟐 𝒙 =
𝟏+𝒄𝒐𝒔 𝟐𝒙
𝟐
(iii) 𝒔𝒊𝒏𝟑𝒙 = 𝟑 𝒔𝒊𝒏 𝒙−𝒔𝒊𝒏 𝟑𝒙
𝟒 (iv)𝒄𝒐𝒔𝟑𝒙 =
𝟑 𝒄𝒐𝒔 𝒙+𝒄𝒐𝒔 𝟑𝒙
𝟒
(v) tan2 x = sec2 x – 1 (vi) cot2 x = cosec2x – 1
(Vii) 2sin x sin y = cos (x – y ) – cos ( x + y)
(vii) 2 cos x cos y = cos (x + y) + cos (x – y )
(ix) 2 sin x cos y = sin (x + y ) + sin (x – y )
(x) 2cos x sin y = sin (x + y ) – sin ( x – y )
2. Integration by substitution
(a) Consider I= dxxf )(
Put x=g(t) so that ).(1 tgdt
dx
We write dx=g1(t). Thus I= dttgtgfdxxf )())(()( 1
(b) When the integrand is the product of two functions and one of them is a function g
(x) and the other is k g’ (x), where k is a constant then
Put g (x) = t
3. Integration of the types ∫𝒅𝒙
𝒂𝒙𝟐+𝒃𝒙+𝒄 ,∫
𝒅𝒙
√𝒂𝒙𝟐+𝒃𝒙+𝒄, and ∫(𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄)𝒅𝒙
In this three forms change 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 in the form A2 + X2 , X2 – A2, or A2 – X2
Where X is of the form x +k and A is a constant ( by completing square method)
Then integral can be find by using any of the special integral formulae
4. Integration of the types ∫𝒑𝒙+𝒒
𝒂𝒙𝟐+𝒃𝒙+𝒄 𝒅𝒙 ,∫
𝒑𝒙+𝒒
√𝒂𝒙𝟐+𝒃𝒙+𝒄 𝒅𝒙 and ∫(𝒑𝒙 +
𝒒)√𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 dx
30
In this three forms split the linear px +q = 𝝀 𝒅
𝒅𝒙(𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 ) + 𝝁
Then divide the integral into two integrals
The first integral can be find out by method of substitution and the second integral
by completing square method as explained in 3
I,e., to evaluate ∫𝒑𝒙+𝒒
√𝒂𝒙𝟐+𝒃𝒙+𝒄 𝒅𝒙 = ∫
𝝀 (𝟐𝒂𝒙+𝒃 )+ 𝝁
𝒂𝒙𝟐+ 𝒃𝒙+𝒄 dx
= 𝝀∫𝟐𝒂𝒙+𝒃
𝒂𝒙𝟐+ 𝒃𝒙+𝒄 𝒅𝒙 + 𝝁∫
𝒅𝒙
𝒂𝒙𝟐+ 𝒃𝒙+𝒄
↓ ↓
Find by substitution method + by completing square
method
5. Integration of rational functions
In the case of rational function, if the degree of the numerator is equal or greater
than degree of the denominator , then first divide the numerator by denominator and
write it as 𝑵𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 = 𝑸𝒖𝒐𝒕𝒊𝒆𝒏𝒕 +
𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓 , then integrate
6. Integration by partial fractions
Integration by partial fraction is applicable for rational functions. There first we must
check thatdegree of the numerator is less than degree of the denominator, if not, divide
the numerator by denominator and write as 𝑵𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 = 𝑸𝒖𝒐𝒕𝒊𝒆𝒏𝒕 +
𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓
and proceed for partial fraction of 𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓
Sl.
No.
Form of the rational functions Form of the rational functions
1 𝒑𝒙 + 𝒒
(𝒙 − 𝒂)(𝒙 − 𝒃 )
𝑨
𝒙 − 𝒂+
𝑩
𝒙 − 𝒃
2 𝒑𝒙 + 𝒒
(𝒙 − 𝒂)𝟐
𝑨
𝒙 − 𝒂+
𝑩
(𝒙 − 𝒂)𝟐
31
3. 𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓
(𝒙 − 𝒂 )(𝒙 − 𝒃 )𝒙 − 𝒄 )
𝑨
𝒙 − 𝒂+
𝑩
𝒙 − 𝒃+
𝑪
𝒙 − 𝒄
4 𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓
(𝒙 − 𝒂 )𝟐(𝒙 − 𝒃 )
𝑨
𝒙 − 𝒂+
𝑩
(𝒙 − 𝒂 )𝟐+
𝑪
𝒙 − 𝒃
5 𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓
(𝒙 − 𝒂 )(𝒙𝟐 + 𝒃𝒙 + 𝒄 )
Where x2 + bx + c cannot be
factorized further
𝑨
𝒙 − 𝒂+
𝑩𝒙 + 𝑪
𝒙𝟐 + 𝒃𝒙 + 𝒄
DEFINITE INTEGRATION
Working Rule for different types of definite integrals
1. Problems in which integral can be found by direct use of standard formula or by
transformation method
Working Rule
(i). Find the indefinite integral without constant c
(ii). Then put the upper limit b in the place of x and lower limit a in the place of x and
subtract the second value from the first. This will be the required definite integral.
2. Problems in which definite integral can be found by substitution method
Working Rule
When definite integral is to be found by substitution then change the lower and upper
limits of integration. If substitution is z = φ(x) and lower limit integration is a and
upper limit is b Then new lower and upper limits will be φ(a) and φ(b) respectively.
Properties of Definite integrals
1. ∫ 𝒇(𝒙)𝒅𝒙 =𝒃
𝒂∫ 𝒇(𝒕)𝒅𝒕
𝒃
𝒂
𝟐. ∫ 𝒇(𝒙)𝒅𝒙 =𝒃
𝒂∫ 𝒇(𝒙)𝒅𝒙
𝒂
𝒃. In particular, ∫ 𝒇(𝒙)𝒅𝒙 =
𝒂
𝒂 0
3. ∫ 𝒇(𝒙)𝒅𝒙 =𝒃
𝒂∫ 𝒇(𝒙)𝒅𝒙
𝒄
𝒂 + ∫ 𝒇(𝒙)𝒅𝒙
𝒃
𝒄, a < c < b
4. ∫ 𝒇(𝒙)𝒅𝒙 =𝒃
𝒂∫ 𝒇(𝒂 + 𝒃 − 𝒙)𝒅𝒙
𝒃
𝒂
5. ∫ 𝒇(𝒙)𝒅𝒙 =𝒂
𝟎∫ 𝒇(𝒂 − 𝒙)𝒅𝒙
𝒂
𝟎
6. ∫ 𝒇(𝒙)𝒅𝒙 =𝟐𝒂
𝟎∫ 𝒇(𝒙)𝒅𝒙
𝒂
𝟎 + ∫ 𝒇(𝟐𝒂 − 𝒙)𝒅𝒙
𝒂
𝟎
7. ∫ 𝒇(𝒙)𝒅𝒙 =𝟐𝒂
𝟎𝟐∫ 𝒇(𝒙)𝒅𝒙
𝒂
𝟎 , if 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙) and
=0 , if 𝒇(𝟐𝒂 − 𝒙) = −𝒇(𝒙)
8. (i) ∫ 𝒇(𝒙)𝒅𝒙 =𝒂
−𝒂𝟐∫ 𝒇(𝒙)𝒅𝒙
𝒂
𝟎, if 𝒇 is an even function, i.e., if 𝒇(-x) = 𝒇(𝒙)
(ii) ∫ 𝒇(𝒙)𝒅𝒙 =𝒂
−𝒂𝟎, if 𝒇 is an odd function, i.e., if 𝒇(−x) = −𝒇(𝒙)
32
Problem based on property
∫ 𝒇(𝒙)𝒅𝒙 =𝒃
𝒂∫ 𝒇(𝒙)𝒅𝒙
𝒄
𝒂 + ∫ 𝒇(𝒙)𝒅𝒙
𝒃
𝒄, a < c < b
Working Rule
This property should be used if the integrand is different in different parts of the
interval [a,b] in which function is to be integrand. This property should also be used
when the integrand (function which is to be integrated) is under modulus sign or is
discontinuous at some points in interval [a,b]. In case integrand contains modulus then
equate the functions whose modulus occur to zero and from this find those values of x
which lie between lower and upper limits of definite integration and then use the
property.
Problem based on property
∫ 𝒇(𝒙)𝒅𝒙 =𝒂
𝟎∫ 𝒇(𝒂 − 𝒙)𝒅𝒙
𝒂
𝟎
Working Rule
Let I = ∫ 𝒇(𝒙)𝒅𝒙𝒂
𝟎
Then I = ∫ 𝒇(𝒂 − 𝒙)𝒅𝒙𝒂
𝟎
(1) + (2) => 2I = ∫ 𝒇(𝒙)𝒅𝒙𝒂
𝟎 + ∫ 𝒇(𝒂 − 𝒙)𝒅𝒙
𝒂
𝟎
I = 𝟏
𝟐∫ {𝒇(𝒙) + 𝒇(𝒂 − 𝒙)}𝒅𝒙
𝒂
𝟎
This property should be used when 𝒇(𝒙) + 𝒇(𝒂 − 𝒙) becomes an integral function of x.
Problem based on property
∫ 𝒇(𝒙)𝒅𝒙 =𝒂
−𝒂 0, if 𝒇(𝒙) is an odd function and ∫ 𝒇(𝒙)𝒅𝒙 =
𝒂
−𝒂 2∫ 𝒇(𝒙)𝒅𝒙
𝒂
𝟎, if 𝒇(𝒙) is an
even function.
Working Rule
This property should be used only when limits are equal and opposite and the
function which is to be integrated is either odd/ even.
PROBLEM BAESD ON LIMIT OF SUM
Working rule
∫ 𝒇 (𝒙)𝒅𝒙 𝒃
𝒂 = 𝐥𝐢𝐦
𝒉→𝟎𝒉 { 𝒇 (𝒂) + 𝒇 (𝒂 + 𝒉 ) + … .+𝒇 ( 𝒂 + (𝒏 − 𝟏 )𝒉 )}
Where nh = b – a
The following results are used for evaluating questions based on limit of sum.
(i) 1 + 2 + 3 + ….. + (n-1) =
1
1 2
)1(n
k
nnk
(ii) 𝟏𝟐 + 𝟐𝟐 + 𝟑𝟐 + …. + (𝒏 − 𝟏)𝟐 = ∑(𝒏−𝟏)𝒏(𝟐𝒏−𝟏)
𝟔
(iii) 𝟏𝟑 + 𝟐𝟑 + 𝟑𝟑 + ….. + (𝒏 − 𝟏)𝟑 = [(𝒏−𝟏)𝒏
𝟐]𝟐
(iv) a + ar + …… + 𝒂𝒓𝒏−𝟏 = 𝒂[𝒓𝒏− 𝟏]
𝒓−𝟏 (r≠1)
IMPORTANT SOLVED PROBLEMS
Evaluate the following integrals
33
1. ∫( 𝟏+𝒍𝒐𝒈 𝑿 )𝟐
𝑿 𝒅𝒙
Solution
put 1+log x = t 𝟏
𝒙 𝒅𝒙 = 𝒅𝒕
∫( 𝟏+𝒍𝒐𝒈 𝑿 )𝟐
𝑿 𝒅𝒙 = ∫ 𝒕𝟐 𝒅𝒕
= 𝒕𝟑
𝟑+ 𝒄
= (𝟏+𝒍𝒐𝒈𝒙 )𝟑
𝟑 + 𝒄
2. ∫𝒆𝒙
√𝟓−𝟒𝒆𝒙−𝒆𝟐𝒙 𝒅𝒙
Solution
Put 𝒆𝒙 = 𝒕 𝒕𝒉𝒆𝒏 𝒆𝒙 𝒅𝒙 = 𝒅𝒕
∫𝒆𝒙
√𝟓 − 𝟒𝒆𝒙 − 𝒆𝟐𝒙 𝒅𝒙 = ∫
𝒅𝒕
√𝟓 − 𝟒𝒕 − 𝒕𝟐
= ∫𝒅𝒕
√− (𝒕𝟐+ 𝟒𝒕−𝟓 )
= ∫𝒅𝒕
√− ( 𝒕𝟐+ 𝟒𝒕+𝟒−𝟒−𝟓 )
= ∫𝒅𝒕
√− { ( 𝒕+𝟐 )𝟐−𝟗 }
= ∫𝒅𝒕
√𝟑𝟐−( 𝒕+𝟐 )𝟐
= 𝐬𝐢𝐧−𝟏 𝒕+𝟐
𝟑 + C = 𝐬𝐢𝐧−𝟏 (
𝒆𝒙+𝟐
𝟑) + 𝑪
3. ∫√𝒕𝒂𝒏𝒙 dx
Solution
Put tan x = t 2then sec2x dx = 2t dt => dx = 𝟐𝒕 𝒅𝒕
𝟏+ 𝒕𝟒
∫√𝒕𝒂𝒏𝒙 dx = ∫ 𝒕 𝟐𝒕 𝒅𝒕
𝟏+ 𝒕𝟒 = ∫
𝟐𝒕𝟐
𝟏+ 𝒕𝟒 𝒅𝒕
= ∫𝟐
𝟏
𝒕𝟐+ 𝒕𝟐
𝒅𝒕 (by dividing nr and dr by t2 )
= ∫(𝟏+
𝟏
𝒕𝟐) + (𝟏−
𝟏
𝒕𝟐)
𝒕𝟐+ 𝟏
𝒕𝟐
𝒅𝒕
= ∫𝟏+
𝟏
𝒕𝟐
𝒕𝟐+ 𝟏
𝒕𝟐
𝒅𝒕 + ∫𝟏 −
𝟏
𝒕𝟐
𝒕𝟐+ 𝟏
𝒕𝟐
𝒅𝒕
= ∫𝟏+
𝟏
𝒕𝟐
(𝒕− 𝟏
𝒕)𝟐+ 𝟐
𝒅𝒕 + ∫𝟏−
𝟏
𝒕𝟐
(𝒕+ 𝟏
𝒕)𝟐− 𝟐
𝒅𝒕
= ∫𝒅𝒖
𝒖𝟐+ 𝟐 + ∫
𝒅𝒗
𝒗𝟐− 𝟐 ( 1st integral put 𝒕 −
𝟏
𝒕 = u
then (𝟏 + 𝟏
𝒕𝟐 ) 𝒅𝒕 = 𝒅𝒖
2nd integral put𝒕 +𝟏
𝒕= 𝒗 𝒕𝒉𝒆𝒏 ( 𝟏 −
𝟏
𝒕𝟐)𝒅𝒕 = 𝒅𝒗
=𝟏
√𝟐𝐭𝐚𝐧−𝟏 (
𝒖
√𝟐) +
𝟏
𝟐√𝟐𝒍𝒐𝒈 |
𝒗− √𝟐
𝒗+√𝟐| + 𝑪
34
= 𝟏
√𝟐𝐭𝐚𝐧−𝟏 (
𝒕−𝟏
𝒕
√𝟐) +
𝟏
𝟐√𝟐𝒍𝒐𝒈 |
𝒕+𝟏
𝒕− √𝟐
𝒕+𝟏
𝒕+√𝟐
| + 𝑪
= 𝟏
√𝟐𝐭𝐚𝐧−𝟏 (
𝒕𝟐− 𝟏
√𝟐 𝒕) +
𝟏
𝟐√𝟐𝒍𝒐𝒈 |
𝒕𝟐+ 𝟏− √𝟐 𝒕
𝒕𝟐+ 𝟏+ √𝟐 𝒕| + 𝑪
= 𝟏
√𝟐𝐭𝐚𝐧−𝟏 (
𝒕𝒂𝒏𝒙− 𝟏
√𝟐 𝒕𝒂𝒏𝒙) +
𝟏
𝟐√𝟐𝒍𝒐𝒈 |
𝒕𝒂𝒏𝒙+ 𝟏− √𝟐𝒕𝒂𝒏𝒙
𝒕𝒂𝒏𝒙+ 𝟏− √𝟐𝒕𝒂𝒏𝒙| + 𝑪
4. ∫𝟓𝒙+𝟑
√𝒙𝟐+ 𝟒𝒙+ 𝟏𝟎 𝒅𝒙
Solution
5x + 3 = A (2x + 4 ) + B = > A = 𝟓
𝟐 𝒂𝒏𝒅 𝑩 = −𝟕
∫𝟓𝒙 + 𝟑
√𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎 𝒅𝒙 = ∫
𝟓
𝟐(𝟐𝒙 + 𝟒) − 𝟕
√𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎 𝒅𝒙
= ∫𝟓
𝟐(𝟐𝒙+𝟒)
√𝒙𝟐+ 𝟒𝒙+𝟏𝟎 𝒅𝒙 + ∫
𝟕
√𝒙𝟐+ 𝟒𝒙+𝟏𝟎𝒅𝒙
= 𝟓
𝟐∫
(𝟐𝒙+𝟒)
√𝒙𝟐+ 𝟒𝒙+𝟏𝟎 𝒅𝒙 + 𝟕 ∫
𝟏
√𝒙𝟐+ 𝟒𝒙+𝟏𝟎𝒅𝒙
= 𝟓
𝟐∫
𝒅𝒕
√𝒕 + 𝟕 ∫
𝟏
√𝒙𝟐+ 𝟒𝒙+𝟒−𝟒+𝟏𝟎𝒅𝒙
= 𝟓
𝟐× 𝟐√𝒕 + 𝟕 ∫
𝟏
√(𝒙+𝟐 )𝟐+𝟔 dx
= 5 √𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎 + 𝟕 𝒍𝒐𝒈 | 𝒙 + 𝟐 + √𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎| + 𝑪
5. ∫𝒙 𝒕𝒂𝒏 𝒙
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙
𝝅
𝟎
Solution
I = ∫𝒙 𝒕𝒂𝒏 𝒙
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙
𝝅
𝟎 -------------------(1)
Again I = ∫(𝝅−𝒙) 𝒕𝒂𝒏(𝝅−𝒙)
𝒔𝒆𝒄 (𝝅−𝒙)+𝒕𝒂𝒏 (𝝅−𝒙) 𝒅𝒙
𝝅
𝟎 Using the property ∫ 𝒇 (𝒙)𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙 )𝒅𝒙
𝒂
𝟎
𝒂
𝟎
I = ∫(𝝅−𝒙) 𝒕𝒂𝒏 𝒙
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙
𝝅
𝟎 -------------------- (2)
Adding (1) and (2) we get
2 I = ∫𝒙 𝒕𝒂𝒏 𝒙
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙
𝝅
𝟎+ ∫
(𝝅−𝒙) 𝒕𝒂𝒏 𝒙
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙
𝝅
𝟎
= 𝝅∫𝒕𝒂𝒏𝒙
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙𝒅𝒙
𝝅
𝟎 = 𝝅∫
𝒕𝒂𝒏𝒙 (𝒔𝒆𝒄 𝒙−𝒕𝒂𝒏 𝒙 )
𝒔𝒆𝒄𝟐𝒙− 𝒕𝒂𝒏𝟐𝒙
𝝅
𝟎 𝒅𝒙
= 𝝅∫ (𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄 𝒙 − 𝒕𝒂𝒏𝟐𝝅
𝟎𝒙) dx = 𝝅 ∫ (𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄 𝒙 − 𝒔𝒆𝒄𝟐𝒙 + 𝟏)𝒅𝒙
𝝅
𝟎
= 𝝅⌈𝒔𝒆𝒄 𝒙 − 𝒕𝒂𝒏 𝒙 + 𝒙⌉𝝅𝟎
I = 𝝅(𝝅
𝟐− 𝟏)
6. Evaluate ∫ (𝒙𝟐𝟒
𝟏− 𝒙) 𝒅𝒙 using limit of sum
Solution
Comparing the given integral with ∫ 𝒇 (𝒙)𝒅𝒙 𝒃
𝒂
a = 1, b = 4 f (x ) = x2 – x
∴ 𝒏𝒉 = 𝟒 − 𝟏 = 𝟑 and
f ( a+ (n-1)h )= f ( 1 + (n – 1 ) h )
= ( 1 + (n – 1 ) h ) 2 - [1 + (n – 1 ) h]
= 1 + 2 (n-1 )h + (n – 1 ) 2 h2 -1 – (n – 1 ) h
35
= (n – 1 ) h + (n – 1 ) 2 h2
∫ 𝒇 (𝒙)𝒅𝒙 𝒃
𝒂 = 𝐥𝐢𝐦
𝒉 →𝟎 𝒉 ∑ 𝒇 ( 𝒂 + (𝒏 − 𝟏 )𝒉 )𝒏−𝟏
𝟎
∫ (𝒙𝟐𝟒
𝟏− 𝒙) 𝒅𝒙 = 𝐥𝐢𝐦
𝒉→𝟎𝒉{∑ (𝒏 − 𝟏 ) 𝒉 + (𝒏 − 𝟏)𝟐𝒏−𝟏
𝟎 𝒉𝟐}
= 𝐥𝐢𝐦𝒉→𝟎
𝒉{∑(𝒏 − 𝟏 )𝒉 + ∑(𝒏 − 𝟏)𝟐𝒉𝟐}
= 𝐥𝐢𝐦𝒉→𝟎
𝒉{𝒉∑(𝒏 − 𝟏 ) + 𝒉𝟐 ∑(𝒏 − 𝟏)𝟐}
= 𝐥𝐢𝐦𝒉→𝟎
𝒉𝟐 𝒏 ( 𝒏−𝟏 )
𝟐 + 𝒉𝟑 𝒏 ( 𝒏−𝟏 )( 𝟐𝒏−𝟏 )
𝟔
= 𝐥𝐢𝐦𝒉 →𝟎
𝒏𝒉 ( 𝒏𝒉−𝒉 )
𝟐+
𝒏𝒉 ( 𝒏𝒉−𝒉 )(𝟐𝒏𝒉−𝒉 )
𝟔
= 𝐥𝐢𝐦𝒉→𝟎
𝟑 ( 𝟑−𝒉 )
𝟐+
𝟑 ( 𝟑−𝒉 )( 𝟔−𝒉 )
𝟔
= 𝟗
𝟐 +
𝟓𝟒
𝟔
= 𝟐𝟕
𝟐
PRACTICE PROBLEMS
LEVEL I
Evaluate the following integrals
1. ∫(𝟐𝒙 − 𝟑 𝒄𝒐𝒔𝒙 + 𝒆𝒙) 𝒅𝒙
2. ∫𝒙𝟒+𝟓𝒙𝟐+ 𝟑
√𝒙 𝒅𝒙
3. ∫ 𝒔𝒊𝒏𝟑 𝒙 𝒅𝒙
4. ∫ 𝒔𝒊𝒏 𝟑𝒙 𝒄𝒐𝒔 𝟓𝒙 𝒅𝒙
5. ∫ 𝒔𝒊𝒏𝟑𝒙𝒄𝒐𝒔𝟓𝒙 𝒅𝒙
6. ∫𝒅𝒙
𝒙𝟐+ 𝟒𝒙+𝟏
7. ∫𝒅𝒙
√𝒙𝟐+ 𝟓𝒙+𝟖
8. ∫ 𝒕𝒂𝒏𝟐𝒙𝝅
𝟒𝟎
𝒅𝒙
9. ∫𝒙𝟐𝒆𝒙 𝒅𝒙
10. ∫𝒙
𝒙𝟐+ 𝟏
𝟏
𝟎 𝒅𝒙
36
36
LEVEL II
Evaluate the following integrals
1. ∫𝒔𝒊𝒏 𝒙
𝒔𝒊𝒏 (𝒙−𝒂 ) 𝒅𝒙
2. ∫𝟐
( 𝟏−𝒙 )( 𝟏+ 𝒙𝟐) 𝒅𝒙
3. ∫𝒙+𝟐
√( 𝒙−𝟓 )(𝒙−𝟑 ) 𝒅𝒙
4. ∫ 𝒆𝒙 (𝒔𝒊𝒏 𝟒𝒙−𝟒
𝟏−𝒄𝒐𝒔 𝟒𝒙) 𝒅𝒙
5. ∫𝒙𝟐 𝐭𝐚𝐧−𝟏 𝒙 𝒅𝒙
6. ∫𝒅𝒙
𝒄𝒐𝒔𝟐𝒙+𝒔𝒊𝒏 𝟐𝒙
7. ∫𝟐𝒙
( 𝟏+ 𝒙𝟐)( 𝟑+ 𝒙𝟐) 𝒅𝒙
9. ∫ 𝐭𝐚𝐧−𝟏 √𝟏−𝒙
𝟏+𝒙 𝒅𝒙
10. ∫𝒙𝒕𝒂𝒏 𝒙
𝒔𝒆𝒄 𝒙 𝒄𝒐𝒔𝒆𝒄 𝒙
𝝅
𝟎 𝒅𝒙
11. ∫𝒙
𝟏+𝒔𝒊𝒏𝒙
𝝅
𝟎 𝒅𝒙
12. ∫ | 𝒙𝟑 − 𝒙 |𝟑
−𝟏 𝒅𝒙
13. ∫√𝒙
√𝒙+ √𝟓−𝒙
𝟑
𝟏 𝒅𝒙
14. ∫ 𝒍𝒐𝒈 ( 𝟏 + 𝒕𝒂𝒏 𝒙 )𝒅𝒙𝝅
𝟒𝟎
15. ∫ (𝒙𝟐 + 𝟐) 𝒅𝒙 𝟐
𝟎 as a limit of sum
16.
2
0
35
dxCosxSinx
CosxSinx
LEVEL III
Evaluate the following integrals
1. ∫𝒔𝒊𝒏 𝒙
𝒔𝒊𝒏𝟒𝒙 𝒅𝒙
2. ∫(√𝒕𝒂𝒏 𝒙 + √𝒄𝒐𝒕 𝒙 ) 𝒅𝒙
3. ∫𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙
𝒔𝒊𝒏𝟒𝒙+ 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙
4. ∫√𝟏− √𝒙
𝟏+√𝒙 𝒅𝒙
5. ∫ 𝒔𝒊𝒏 (𝒍𝒐𝒈 𝒙 )𝒅𝒙
6. ∫ 𝒍𝒐𝒈 ( 𝒙 + √𝒙𝟐 + 𝒂𝟐) 𝒅𝒙
7 ∫ 𝒙 𝒍𝒐𝒈 (𝒔𝒊𝒏 𝒙 )𝒅𝒙𝝅
𝟎
8. ∫ (𝟐𝒙𝟐 + 𝒙 + 𝟕 ) 𝒅𝒙𝟐
𝟏 as a limit of sum
37
37
9. xx
xdx
tan1
10 dx
xx
x
)1)(1( 2
4
11.
dxx
x
2log
1loglog