Question

Find the integral.

$\int {\tan }^{2}4xsec4xdx$

Short answer

Answer is$\frac{-1}{8}\ln \left(\left|\tan \left(2x+\frac{\mathrm{\pi }}{4}\right)\right|\right)+\frac{1}{8}\tan 4xsec4x+C$

Step-by-step solution

Step 1. Given information

Integral is$\int {\tan }^{2}4xsec4xdx$

Step 2. Explanation

Let$4x=u$

$4dx=du$

$$\begin{gathered} \int {\tan }^{2}4xsec4xdx=\frac{1}{4}\int {\tan }^{2}usecudu \\ =\frac{1}{4}\int \left(se{c}^{2}u-1\right)secudu \\ =\frac{1}{4}\int se{c}^{3}u-secudu \\ =\frac{-1}{8}\ln \left(\left|\tan \left(\frac{u}{2}+\frac{\mathrm{\pi }}{4}\right)\right|\right)+\frac{1}{8}\tan usecu+C \\ =\frac{-1}{8}\ln \left(\left|\tan \left(2x+\frac{\mathrm{\pi }}{4}\right)\right|\right)+\frac{1}{8}\tan 4xsec4x+C \end{gathered}$$