RD Sharma Solutions Class 12 Maths Chapter 19 Ex 19 · Indefinite Integrals Ex 19.9 Q1 Let I= J logx dx X Let logx = t then, d (logx) = dt 1 => -dx = dt X => dx = xdt Putting logx

RD Sharma

Solutions

Class 12 Maths

Chapter 19Ex 19.8

Indefinite Integrals Ex 19.9 Q1

Let I = J logx

dxX

Let logx = t then, d (logx) = dt

1 => -dx = dt

X

=> dx = xdt

Putting logx = t and dx = x dt, we get

t l=J-xxdt

X

= J tdt

t2 =-+C

2

(1ogx)2

=---+C

2

(logx )2

I = -'---------'- + c 2
















































Let

Let

sin ( tan-1 x) I= J--'------'-dX ---- -(i)

l+x2

tan-1 x = t then,

d (tan-1 x) = dt

1 --dx=dt l+x2

Putting tan-1 x = t and � = dt in equation (i) ,l+x2

we get

I= Jsintdt

= - COS t+ C

= -cos(tan-1 x) +c

I= -cos (tan-1 x) +c


Let I = J sin (lo gx)

dx -----(i) X

Let logx = t then, d (logx) = dt

1 -dx =dtX

1Putting logx = t and -dx = dt in equation (i), X

we get

I= Jsintdt

=-cost+c = - cos (logx) +c

I= -cos(logx) +c

















Let

Let

sin5 x I= J--dx----(1) cos4 x

cosx = t d (cosx) = dt

-sinxdx =dtdt dx = --

sinx

then,

Putting

we get

dt cosx = t and dx = - -- in equation (i),sinx

l = J si n5

X

X _ �t4 sinx

. 4

= -J sin x dt

t4

(1- cos2 x)2

= -J

t4 dt

2

( 1 - t2) = -J

t4 dt

4 2

= -J 1 + t - 2t dt

t4

= -J (_:_ + t4

-

2t2 )dt

t4 t4 t4

= -J ( t-4 + 1 - 2t-2 ) dt

=- -+t-2- +c[ t-3 t-1]

-3 -1

= -[-2-x2-+ t+ �]+c3 t3 t

1 1 2 =-x--t--+c 3 t3 t

1 1 2 = -x-....,,...-- COSX - --+C

3 cos3 x cosx

2 1 l = -COSX---+---+C

cosx 3cos3 x

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RD Sharma Solutions Class 12 Maths Chapter 19 Ex 19 · Indefinite Integrals Ex 19.9 Q1 Let I= J logx dx X Let logx = t then, d (logx) = dt 1 => -dx = dt X => dx = xdt Putting logx