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Question: Answered & Verified by Expert
$\int \frac{1}{\cos x \sqrt{\cos 2 x}} d x=$
(where C is a constant of integration.)
MathematicsIndefinite IntegrationMHT CETMHT CET 2022 (06 Aug Shift 1)
Options:
  • A $\sin ^{-1}(\tan x)+C$
  • B $\log \left(\tan x+\sqrt{\tan ^2 x+1}\right)+C$
  • C $\tan ^{-1} x+C$
  • D $\log \left(\tan x+\sqrt{\tan ^2 x-1}\right)+C$
Solution:
1004 Upvotes Verified Answer
The correct answer is: $\sin ^{-1}(\tan x)+C$
$\begin{aligned} & \int \frac{d x}{\cos x \sqrt{\cos 2 x}}=\int \frac{d x}{\cos x \sqrt{\frac{1-\tan ^2 x}{1+\tan ^2 x}}} \\ & \int \frac{d x}{\cos x \frac{\sqrt{1-\tan ^2 x}}{\sec x}}=\int \frac{\sec ^2 d x}{\sqrt{1-\tan ^2 x}} \\ & \int \frac{d t}{\sqrt{1-t^2}}=\sin ^{-1}(t)+C=\sin ^{-1}(\tan x)+C\end{aligned}$
[let $\tan \mathrm{x}=\mathrm{t}$ ]

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