$\int \frac{d x}{\cos x \sqrt{\cos 2 x}}=$

$\int \frac{d x}{\cos x \sqrt{\cos 2 x}}=$
  1. $\sin ^{-1}(\tan x)+c$
  2. $\frac{1}{2} \log \left|\tan \left(\frac{\pi}{4}+\mathrm{x}\right)\right|+\mathrm{c}$
  3. $2 \log \left|\frac{1+\tan x}{1-\tan x}\right|+c$
  4. $\frac{1}{2} \log \left|\frac{1-\tan x}{1+\tan x}\right|+c$

Solution

$\begin{aligned} & \text { Let } I=\int \frac{d x}{\cos x \sqrt{\cos 2 x}} \\ & =\int \frac{d x}{\cos x \cdot \cos x \sqrt{\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}}}=\int \frac{d x}{\cos ^2 x \sqrt{1-\tan ^2 x}} \\ & I=\int \frac{\sec ^2 x}{\sqrt{1-\tan ^2 x}} d x \end{aligned}$ Put $\tan x=t \Rightarrow \sec ^2 d x=d t$ $\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{\sqrt{1-\mathrm{t}^2}}=\sin ^{-1}(\mathrm{t})+\mathrm{c}=\sin ^{-1}(\tan \mathrm{x})+\mathrm{c}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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