Question

How do you integrate `int xtan^2x` using integration by parts?

Answer 1

Use `tan^2 x = sec^2 x -1` first.

Explanation:

`x tan^2 x = xsec^2 x - x`

`-int x dx = -x^2/2`

`int x sec^2 x dx`

Let `u = x` and `dv = sec^2 x dx`, so that

`du = dx` and `v = tan x` to get

`int x sec^2 x dx = x tan x - int tan x dx`.

Now integrate `tan x = sinx/cosx` using substitution `u = cos x`.

`int x sec^2 x dx = x tan x - (- ln abs(cosx)) = x tan x + ln abs cosx`

Finish by putting it all together.

`int x tan^2 x = -x^2/2 + x tan x + ln abs cosx +C`.