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

2.  4sin(2x + ) dx  cos(4 - ) d  -6sin(3 + /2) d Example
= 1/4sin(4 - ) + C
Example
 -6sin(3 + /2) d
= 1/3 X 6cos(3 + /2) + C
= 2cos(3 + /2) + C
Example 
 4sin(2x + ) dx
= [ ] 
½ X –4cos(2x + )
/2
/2
= [ ] 
–2cos(2x + )
/2
= (-2cos3) - (-2cos2)
= 2 – (-2)
= 4

3. Example
By firstly rearranging the formula cos2 = 2cos2 - 1
Find  cos2 d
/2
***********
 cos2 d
/2
cos2 = 2cos2 - 1
/2
2cos2 - 1 = cos2
=  (1/2cos2 + 1/2) d
2cos2 = cos2 + 1
= [ ½ X 1/2sin2 + 1/2 ]
/2
cos2 = 1/2cos2 + 1/2
= [1/4sin2 + 1/2 ]
/2
= (/4 + 1/4sin) - ( sin0)
= /4

4. Example Find the shaded area !! Curves cross when sin2x = cosx
y = sin2x B
y = cosx C
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 B
A C

5. First area =  (sin2x – cosx) dx
/2
/6
= [ ]
/2
-1/2cos2x - sinx
/6
= ( -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