Mathematics | MHT CET 2018 | Discussion Page

$\int_{0}^{\pi/4} x.\sec^{2}x dx=$

$\frac{\pi}{4}+\log \sqrt{2}$

$\frac{\pi}{4}-\log \sqrt{2}$

$1+\log \sqrt{2}$

$1-\frac{1}{2}\log {2}$

Option B

Let l= $\int_{0}^{\pi/4} x.\sec^{2}x dx$

$\Rightarrow$ $l=[x\int \sec^{2}x dx]^{\pi/4}_{0}-\int_{0}^{\pi/4} \left(\frac{dy}{dx}\int \sec^{2}x dx\right) dx$

[integration by parts]

$\Rightarrow$ $l=[x\tan^{}x ]^{\pi/4}_{0}-\int_{0}^{\pi/4} \tan x dx$

$\Rightarrow$ $l=[\frac{\pi}{4}\tan^{}\frac{\pi}{4}-0 ]-[\log \sec x]^{\pi/4}_{0}$

$\Rightarrow$ $l=\frac{\pi}{4}-\left( \log \sec\frac{\pi}{4}-\log \sec 0\right)$

$l=\frac{\pi}{4}-\left( \log\sqrt{2}-\log 1\right)$

$\Rightarrow$ l= $\frac{\pi}{4}-\log \sqrt{2}$

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