Question

How do you find the antiderivative of `int xsec^2(x^2)tan(x^2)dx` from `[0,sqrtpi/2]`?

Answer 1

`1/4`

Explanation:

First, let `u=x^2`. This implies that `du=2xdx`. When making this substitution, remember to plug the current into `u=x^2`. The bound of `0` stays `0` and `sqrtpi/2` becomes `(sqrtpi/2)^2=pi/4`.

Thus:

`int_0^(sqrtpi/2)xsec^2(x^2)tan(x^2)dx=1/2int_0^(pi/4)sec^2(u)tan(u)du`

Here notice that the derivative of tangent is present alongside the tangent function. Let `v=tan(u)` so `dv=sec^2(u)du`. The bounds become `0rarrtan(0)=0` and `pi/4rarrtan(pi/4)=1`.

`=1/2int_0^1vdv=1/2[v^2/2]_0^1=1/2(1^2/2-0^2/2)=1/2(1/2)=1/4`