How do you find the integral of #1/Sin^2 (x) + cos(2x)#?

1 Answer
Jan 4, 2018

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

Explanation:

First, I will split up the integral into two parts:
#int\ 1/sin^2(x)+cos(2x)\ dx=int\ 1/sin^2(x)\ dx + int\ cos(2x)\ dx#

I will call the left one Integral 1 and the right one Integral 2

Integral 1
We can use the following trigonometric identity:
#1/sin(theta)=csc(theta)#

This gives:
#int\ 1/sin^2(x)\ dx=int\ csc^2(x)\ dx#

The derivative of #cot(x)# is #-csc^2(x)#, so we can deduce that the answer to this integral must be #-cot(x)#:
#int\ csc^2(x)\ dx=-cot(x)+C#

Integral 2
Here I will do a u-substitution with #u=2x#. The derivative of #u# is #2#, so we divide by #2# to integrate with respect to #u#:
#int\ cos(2x)\ dx=1/2int\ cos(u)\ du=1/2sin(u)+C#

Undoing the substitution, we get:
#1/2sin(u)+C=1/2sin(2x)+C#

Completing the original integral
Now that we know Integral 1 and Integral 2, we can complete teh original integral:
#int\ 1/sin^2(x)+cos(2x)\ dx=1/2sin(2x)-cot(x)+C#