#int cot (x) cos^2 (x) dx#?

1 Answer
Mar 12, 2018

#int cot (x) cos^2 (x) dx= ln|sin(x)|+ 1/4cos(2x)+C#

Explanation:

Given: #int cot (x) cos^2 (x) dx#

Substitute #cos^2(x) = 1 - sin^2(x)#:

#int cot (x) cos^2 (x) dx= int cot(x)(1-sin^2(x))dx#

Distribute the #cot(x)#:

#int cot (x) cos^2 (x) dx= int cot(x)dx -int cot(x)sin^2(x)dx#

Use the identity #cot(x) = cos(x)/sin(x)#

#int cot (x) cos^2 (x) dx= int cot(x)dx -int cos(x)/sin(x)sin^2(x)dx#

Cancel #sin(x)/sin(x)#:

#int cot (x) cos^2 (x) dx= int cot(x)dx -int cos(x)sin(x)dx#

Use the identity #2cos(x)sin(x) = sin(2x)#

#int cot (x) cos^2 (x) dx= int cot(x)dx - 1/2int sin(2x)dx#

You can find these integrals in any list of integrals:

#int cot (x) cos^2 (x) dx= ln|sin(x)|+ 1/4cos(2x)+C#