Integration of sin(ax + b) & cos(ax + b)

Consider

If f(x) = 1/asin(ax + b)

then f´(x) = 1/acos(ax + b) X a = cos(ax + b)

If g(x) = -1/acos(ax + b)

then g´(x) = 1/asin(ax + b) X a = sin(ax + b)

It now follows that

cos(ax + b)dx = 1/asin(ax + b) + C

sin(ax + b)dx = -1/acos(ax + b) + C

(supplied)

Example

cos(4 - ) d

= 1/4sin(4 - ) + C

Example

-6sin(3 + /2) d= 1/3 X 6cos(3 + /2) + C = 2cos(3 + /2) + C

Example

4sin(2x + ) dx/2

= [ ]½ X –4cos(2x + ) /2

= [ ]–2cos(2x + ) /2

= (-2cos3) - (-2cos2)

= 2 – (-2)

= 4

Example

By firstly rearranging the formula cos2 = 2cos2 - 1

Find cos2 d 0

/2

***********

cos2 = 2cos2 - 1

2cos2 - 1 = cos2

2cos2 = cos2 + 1

cos2 = 1/2cos2 + 1/2

cos2 d0

/2

= (1/2cos2 + 1/2) d0

/2

= [ ½ X 1/2sin2 + 1/2 ] 0/2

= [1/4sin2 + 1/2 ] 0

/2

= (/4 + 1/4sin) - (0 + sin0)

= /4

Example Find the shaded area !!

y = sin2x

y = cosx

Curves cross when sin2x = cosx

2sinxcosx – cosx = 0

cosx(2sinx – 1) = 0

cosx = 0 or sin x = 1/2

x = /2 or 3/2 x = /6 or

5/6

A

B

C

B A C

First area = (sin2x – cosx) dx/2

/6

= [ ] -1/2cos2x - sinx/6

/2

= ( -1/2cos - sin/2) - (-1/2cos /3 - sin/6)

= (1/2 – 1) - (-1/4 – ½)

= ½ - 1+ ¼ + ½

= 1/4

The diagram has ½ turn symmetry about point B

So the total area = 2 X ¼ = 1/2unit2