# How do you integrate int xtan(x^2)sec(x^2) using substitution?

Oct 21, 2016

$\frac{1}{2} \sec \left({x}^{2}\right) + C$

#### Explanation:

$I = \int x \tan \left({x}^{2}\right) \sec \left({x}^{2}\right) \mathrm{dx}$

The first substitution we will make is $u = {x}^{2}$. Differentiating both sides gives us $\frac{\mathrm{du}}{\mathrm{dx}} = 2 x$, so $\mathrm{du} = 2 x \mathrm{dx}$. Notice that in our integrand we currently have $x \mathrm{dx}$, so multiply the interior by $2$ and the exterior by $\frac{1}{2}$ to balance ourselves out.

$I = \frac{1}{2} \int 2 x \tan \left({x}^{2}\right) \sec \left({x}^{2}\right) \mathrm{dx}$

$I = \frac{1}{2} \int \tan \left({x}^{2}\right) \sec \left({x}^{2}\right) \left(2 x \mathrm{dx}\right)$

Substituting in our values for $u$ and $\mathrm{du}$:

$I = \frac{1}{2} \int \tan \left(u\right) \sec \left(u\right) \mathrm{du}$

This is the integral for $\sec \left(u\right)$, since $\frac{d}{\mathrm{du}} \sec \left(u\right) = \sec \left(u\right) \tan \left(u\right)$.

$I = \frac{1}{2} \sec \left(u\right) + C$

Since $u = {x}^{2}$:

$I = \frac{1}{2} \sec \left({x}^{2}\right) + C$