Question

How do you evaluate the integral `int xsec^2x`?

Answer 1

`intxsec^2(x)dx=xtan(x)+ln(abscos(x))+C`

Explanation:

This is a prime candidate for integration by parts, which takes the form `intudv=uv-intvdu`.

For the given integral `intxsec^2(x)dx`, we want to choose a value of `u` that gets simpler when we differentiate it and a value of `dv` that is easily integrated.

So, let:

`{(u=x" "=>" "du=dx),(dv=sec^2(x)dx" "=>" "v=tan(x)):}`

We then have:

`intxsec^2(x)dx=uv-intvdu`

`color(white)(intxsec^2(x)dx)=xtan(x)-inttan(x)dx`

You may have the integral of `tan(x)` memorized. If not, it's easy to find:

`color(white)(intxsec^2(x)dx)=xtan(x)-intsin(x)/cos(x)dx`

Let `t=cos(x)`, implying that `dt=-sin(x)dx`:

`color(white)(intxsec^2(x)dx)=xtan(x)+int(-sin(x))/cos(x)dx`

`color(white)(intxsec^2(x)dx)=xtan(x)+int1/tdt`

This is a common integral:

`color(white)(intxsec^2(x)dx)=xtan(x)+ln(abst)+C`

Working back from `t=cos(x)`:

`color(white)(intxsec^2(x)dx)=xtan(x)+ln(abscos(x))+C`