Question

What is `int sin^4xdx`?

Answer 1

We know that

`sin^2x=1/2(1-cos2x)` and `cos^2 2x=1/2*(1+cos4x)`

Hence

`int sin^4xdx=int (1/2(1-cos2x))^2dx= 1/4*int(1-2cos2x+cos^2 2x)dx= 1/4*int1dx-2*1/4intcos2xdx+1/4intcos^2 2xdx= 1/4*x-1/4*sin2x+1/4*int((1+cos4x)/2)dx= 1/4*x-1/4*sin2x+1/8*x+1/32*sin(4x)+c`