# How would you find this integral using u substitution?

## ${\int}_{- \frac{\pi}{6}}^{\frac{\pi}{6}} {\sec}^{2} \left(2 x\right) \mathrm{dx}$ When I use the U substitution and change x values into u values my answer was -$\sqrt{3}$ I know the answer should be $\sqrt{3}$.

Feb 1, 2018

See explanation.

#### Explanation:

Let $u = 2 x$.

$\mathrm{du} = 2 \mathrm{dx} \rightarrow \mathrm{dx} = \frac{1}{2} \mathrm{du}$

Changing the bounds:

$x = - \frac{\pi}{6} \rightarrow u = 2 \left(- \frac{\pi}{6}\right) = - \frac{\pi}{3}$
$x = \frac{\pi}{6} \rightarrow u = 2 \left(\frac{\pi}{6}\right) = \frac{\pi}{3}$

the new integral becomes:

$\frac{1}{2} {\int}_{- \frac{\pi}{3}}^{\frac{\pi}{3}} {\sec}^{2} \left(u\right) \mathrm{du}$

$= {\left[\frac{1}{2} \tan \left(u\right)\right]}_{- \frac{\pi}{3}}^{\frac{\pi}{3}}$
$= \frac{1}{2} \tan \left(\frac{\pi}{3}\right) - \frac{1}{2} \tan \left(- \frac{\pi}{3}\right)$
$= \frac{1}{2} \sqrt{3} - \frac{1}{2} \left(- \sqrt{3}\right)$
$= \sqrt{3}$