सिद्ध कीजिए _ 0 ^ 4 2 ^3 x dx = 1 - 2

Question

सिद्ध कीजिए $\int_{0}^{\frac\pi 4} 2 \tan^3 x dx = 1 - \log 2$

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Answer

माना $I = \int_{0}^{\frac\pi 4}2 \tan^3x dx = 2 \int_{0}^{\frac\pi 4} \tan^2x \tan x dx$
$= 2 \int_{0}^{\frac\pi 4} (\sec^2 x - 1) \tan x dx (\because 1 + \tan^2 x = \sec^2 x)$
$= 2 [\int_{0}^{\frac\pi 4} \sec^2x \tan x dx - \int_{0}^{\frac\pi 4} \tan x dx]$
$= 2 [\int_{0}^{\frac\pi 4} (\tan x) \sec^2 x dx - 2 [-\log| \cos x]_{0}^{\pi / 4} [\because$
माना $ I_1 = \int(\tan x) \sec^2 x dx, \tan x = t$ रखने पर, $\Rightarrow \sec^2 x dx = dt \therefore I_1 = \int t d t=\frac{t^{2}}{2}=\frac{\tan ^{2} x}{2}]$
$=2\left[\frac{\tan ^{2} x}{2}\right]_{0}^{\frac\pi 4} +2\left[\log \left|\cos \frac{\pi}{4}\right|-\log |\cos 0|\right]$
$=\tan ^{2}\left(\frac{\pi}{4}\right)-0+2 \log \left(\frac{1}{\sqrt{2}}\right)-\log 1 =1+2 \log 2^{-\frac1 2}-0 (\because \log 1 = 0)$
$= 1 - 2 \times \frac{1}{2}\log 2 = 1 - \log 2$

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