Question

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

Answer 1

`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`