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$\int \frac{1}{\cos x \sqrt{\cos 2 x}} d x=$
(where C is a constant of integration.)
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(where C is a constant of integration.)
Solution:
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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 $\tan \mathrm{x}=\mathrm{t}$ ]
[let $\tan \mathrm{x}=\mathrm{t}$ ]
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