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Question

Mathematics Question on integral

∫0π/8tan⁡2(2x)dx=\int\limits_0^{\pi /8} \tan^2 (2x) dx =

A

4−π8\frac{4 - \pi}{8}

B

4+π8\frac{4 + \pi}{8}

C

4−π4\frac{4 - \pi}{4}

D

4−π2\frac{4 - \pi}{2}

Answer

4−π8\frac{4 - \pi}{8}

Explanation

Solution

Let I=∫0π8tan⁡2(2x)dxI = \int_{0}^{\frac{\pi}{8}}\tan^{2} \left(2x\right)dx
I=∫0π8[sec⁡2(2x)−1]dx=[tan⁡2x2−x]0π8I = \int_{0}^{\frac{\pi }{8}}\left[\sec^{2} \left(2x\right) -1 \right]dx =\left[\frac{\tan2x}{2} -x\right]^{\frac{\pi}{8}}_{0}
=[tan⁡π42−π8−tan⁡02+0]=12−π8=4−π8= \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}