Question

Solve the following integral.

$\int se{c}^{3}2xdx$

Short answer

Answer is$\frac{\tan 2xsec2x}{4}+\frac{\ln \left[\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}+x}\right)\right]}{4}+C$

Step-by-step solution

Step 1. Given information

An integral is$\int se{c}^{3}2xdx$

Step 2. Explanation

Let $u=2x$

$\begin{array}{rcl}\int se{c}^{3}2xdx& =& \frac{1}{2}\int se{c}^{3}udu\\ & =& \frac{1}{2}\left[secu\tan u-\int \tan u\tan usecudu\right]\\ & =& \frac{\tan usecu}{4}+\frac{\ln \left[\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}+{\displaystyle \frac{u}{2}}\right)\right]}{4}+C\\ & =& \frac{\tan 2xsec2x}{4}+\frac{\ln \left[\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}+x}\right)\right]}{4}+C\end{array}$