# How do you integrate int sqrttanxsec^2xdx?

Nov 29, 2016

Let $u = \tan \left(x\right)$, then $\mathrm{du} = {\sec}^{2} \left(x\right) \mathrm{dx}$. Please see the explanation.

#### Explanation:

Let $u = \tan \left(x\right)$, then $\mathrm{du} = {\sec}^{2} \left(x\right) \mathrm{dx}$

$\int \sqrt{\tan \left(t\right)} {\sec}^{2} \mathrm{dx} = \int \left({u}^{\frac{1}{2}}\right) \mathrm{du}$

Integrate:

$\int \sqrt{\tan \left(t\right)} {\sec}^{2} \mathrm{dx} = \frac{2}{3} {u}^{\frac{3}{2}} + C$

Reverse the substitution:

$\int \sqrt{\tan \left(t\right)} {\sec}^{2} \mathrm{dx} = \frac{2}{3} {\left(\tan \left(x\right)\right)}^{\frac{3}{2}} + C$