_0^ / 8 ^2(2 x) d x is equal to - Vidyadip

MCQ

$\int_0^{\pi / 8} \tan ^2(2 x) d x$ is equal to

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

✓

Answer

$\begin{array}{l}\text {Let } I=\int_0^{\pi / 8} \tan ^2(2 x) d x=\int_0^{\pi / 8}\left(\sec ^2(2 x)-1\right) d x \\ =\left(\frac{1}{2} \tan 2 x-x\right)_0^{\pi / 8}=\frac{1}{2} \tan 2\left(\frac{\pi}{8}\right)-\frac{\pi}{8}=\frac{1}{2} \tan \frac{\pi}{4}-\frac{\pi}{8} \\ =\frac{1}{2}-\frac{\pi}{8}=\frac{4-\pi}{8}\end{array}$

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