What is the integral of sin x cos^2 2x?

1 Answer
Mar 27, 2018

#intsinxcos^2(2x)dx=-1/2cosx-1/20cos(5x)+1/12cos(3x)+C#

Explanation:

Recall the identity

#cos^2(x)=1/2(1+cos(2x))#

Then, let's temporarily let #u=2x#

#cos^2(2x)=cos^2(u)=1/2(1+cos(2u))=1/2(1+cos(4x))#

Rewrite #intsinxcos^2(2x)dx:#

#1/2intsinx(1+cos(4x))=1/2int(sinx+sinxcos(4x))dx=1/2intsinxdx+1/2intsinxcos(4x)dx=-1/2cosx+1/2intsinxcos4xdx#

Recall the identity

#sinucosv=1/2(sin(u+v)+sin(u-v))#

Then,

#sinxcos4x=1/2(sin(5x)+sin(-3x))=1/2(sin(5x)-sin(3x))#

So, #1/2intsinxcos4xdx=1/4int(sin(5x)-sin(3x))=1/4(-1/5cos(5x)+1/3cos(3x))=-1/20cos(5x)+1/12cos(3x)#

Combining everything together,

#intsinxcos^2(2x)dx=-1/2cosx-1/20cos(5x)+1/12cos(3x)+C#

Adding in the constant of integration.