Question

$\underset{0}{\overset{\pi /8}{\int }}{\tan }^2(2x)dx=$

• 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 $I={\int }_0^{\frac{\pi }{8}}{\tan }^2\left(2x\right)dx$
$I={\int }_0^{\frac{\pi }{8}}\left[{\sec }^2\left(2x\right)-1\right]dx={\left[\frac{\tan 2x}{2}-x\right]}_0^{\frac{\pi }{8}}$
$=\left[\frac{\tan \frac{\pi }{4}}{2}-\frac{\pi }{8}-\frac{\tan 0}{2}+0\right]=\frac{1}{2}-\frac{\pi }{8}=\frac{4-\pi }{8}$