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

Oct 21, 2016

$\frac{1}{4}$

#### Explanation:

First, let $u = {x}^{2}$. This implies that $\mathrm{du} = 2 x \mathrm{dx}$. When making this substitution, remember to plug the current into $u = {x}^{2}$. The bound of $0$ stays $0$ and $\frac{\sqrt{\pi}}{2}$ becomes ${\left(\frac{\sqrt{\pi}}{2}\right)}^{2} = \frac{\pi}{4}$.

Thus:

${\int}_{0}^{\frac{\sqrt{\pi}}{2}} x {\sec}^{2} \left({x}^{2}\right) \tan \left({x}^{2}\right) \mathrm{dx} = \frac{1}{2} {\int}_{0}^{\frac{\pi}{4}} {\sec}^{2} \left(u\right) \tan \left(u\right) \mathrm{du}$

Here notice that the derivative of tangent is present alongside the tangent function. Let $v = \tan \left(u\right)$ so $\mathrm{dv} = {\sec}^{2} \left(u\right) \mathrm{du}$. The bounds become $0 \rightarrow \tan \left(0\right) = 0$ and $\frac{\pi}{4} \rightarrow \tan \left(\frac{\pi}{4}\right) = 1$.

$= \frac{1}{2} {\int}_{0}^{1} v \mathrm{dv} = \frac{1}{2} {\left[{v}^{2} / 2\right]}_{0}^{1} = \frac{1}{2} \left({1}^{2} / 2 - {0}^{2} / 2\right) = \frac{1}{2} \left(\frac{1}{2}\right) = \frac{1}{4}$