Question

What is the integral of `(cosx)^2`?

Answer 1

`1/4sin(2x)+1/2x+C`

Explanation:

We will use the cosine double-angle identity in order to rewrite `cos^2x`. (Note that `cos^2x=(cosx)^2`, they are different ways of writing the same thing.)

`cos(2x)=2cos^2x-1`

This can be solved for `cos^2x`:

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

Thus,

`intcos^2xdx=int(cos(2x)+1)/2dx`

Split up the integral:

`=1/2intcos(2x)dx+1/2intdx`

The second integral is the "perfect integral:" `intdx=x+C`.

`=1/2intcos(2x)dx+1/2x`

The constant of integration will be added upon evaluating the remaining integral.

For the cosine integral, use substitution. Let `u=2x`, implying that `du=2dx`.

Multiply the integrand `2` and the exterior of the integral by `1/2`.

`=1/4int2cos(2x)dx+1/2x`

Substitute in `u` and `du`:

`=1/4intcos(u)du+1/2x`

Note that `intcos(u)du=sin(u)+C`.

`=1/4sin(u)+1/2x+C`

Since `u=2x`:

`=1/4sin(2x)+1/2x+C`

Note that this can be many different ways, since `sin(2x)=2sinxcosx`.