Question

`intsinxcos3xdx`?

Answer 1

`intsin(x)cos(3x)dx=-cos(4x)/8+cos(2x)/4+C`

Explanation:

When the value for `a` in `cos(ax)` is bigger than the value for the value for `b` in `sin(bx)` we say that `sin(a+b)-sin(a-b)=2sin(bx)cos(ax)`

In this case `b=1` and `a=4`, so `sin(x)cos(3x)=(sin(4x)-sin(2x))/2`

`intsin(x)cos(3x)dx=int(sin(4x)-sin(2x))/2dx=intsin(4x)/2-sin(2x)/2dx`

`intsin(ax)dx=-cos(ax)/a+C`

`intsin(4x)/2-sin(2x)/2dx=1/2(-cos(4x)/2+cos(2x)/2)+C=-cos(4x)/8+cos(2x)/4+C`