Question

$ \int\limits_0^{\pi /8} \tan^2 (2x) dx = $

• A. $ \frac{4 - \pi}{8} $
• B. $ \frac{4 + \pi}{8} $
• C. $ \frac{4 - \pi}{4} $
• D. $ \frac{4 - \pi}{2} $

Answer 1

Answer: (A)

Let $I = \int_{0}^{\frac{\pi}{8}}\tan^{2} \left(2x\right)dx $
$I = \int_{0}^{\frac{\pi }{8}}\left[\sec^{2} \left(2x\right) -1 \right]dx =\left[\frac{\tan2x}{2} -x\right]^{\frac{\pi}{8}}_{0}$
$ = \left[ \frac{\tan \frac{\pi}{4}}{2} - \frac{\pi}{8} - \frac{\tan 0}{2} +0\right]= \frac{1}{2}-\frac{\pi}{8}= \frac{4-\pi}{8} $