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$\int \frac{d x}{\cos x \sqrt{\cos 2 x}}=$
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1476 Upvotes
Verified Answer
The correct answer is:
$\sin ^{-1}(\tan x)+c$
$$
\begin{aligned}
& \text { Let } I=\int \frac{d x}{\cos x \sqrt{\cos 2 x}} \\
& =\int \frac{d x}{\cos x \cdot \cos x \sqrt{\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}}}=\int \frac{d x}{\cos ^2 x \sqrt{1-\tan ^2 x}} \\
& I=\int \frac{\sec ^2 x}{\sqrt{1-\tan ^2 x}} d x
\end{aligned}
$$
Put $\tan x=t \Rightarrow \sec ^2 d x=d t$
$$
\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{\sqrt{1-\mathrm{t}^2}}=\sin ^{-1}(\mathrm{t})+\mathrm{c}=\sin ^{-1}(\tan \mathrm{x})+\mathrm{c}
$$
\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}
$$
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