Question

How do you integrate `int tan^2xsec^2x`?

Answer 1

`1/3tan^3x+C`

Explanation:

Look for a function and its derivative inside the integrand. Here, it's very useful to know that the derivative of `tanx` is `sec^2x`.

Here, we have the `tanx` function squared times the derivative of `tanx`.

So, let `u=tanx`, implying that `du=(sec^2x)dx`. Then:

`I=intunderbrace(tan^2x)_(u^2)overbrace((sec^2x)dx)^(du)=intu^2du`

Which is a much simpler problem:

`I=1/3u^3+C=1/3tan^3x+C`