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$\int \frac{\sec x d x}{\sqrt{\cos 2 x}}=$
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Verified Answer
The correct answer is:
$\sin ^{-1}(\tan x)$
$\begin{aligned}
& \int \frac{\sec x d x}{\sqrt{\cos 2 x}}=\int \frac{\sec x}{\sqrt{\cos ^2 x-\sin ^2 x}} d x \\
& \left.=\int \frac{\sec ^2 x d x}{\sqrt{1-\tan ^2 x}} \text { \{Multiplying } N^{\prime} r \text { and } D^{\prime} r \text { by } \sec x\right\}
\end{aligned}$
Now putting $\tan x=t \Rightarrow \sec ^2 x d x=d t$, we get the integral $=\sin ^{-1} t=\sin ^{-1}(\tan x)$.
Trick : Since $\frac{d}{d x}\left\{\sin ^{-1}(\tan x)\right\}=\frac{\sec ^2 x}{\sqrt{1-\tan ^2 x}}$
$=\frac{\sec ^2 x \cdot \cos x}{\sqrt{\cos ^2 x-\sin ^2 x}}=\frac{\sec x}{\sqrt{\cos 2 x}} .$
& \int \frac{\sec x d x}{\sqrt{\cos 2 x}}=\int \frac{\sec x}{\sqrt{\cos ^2 x-\sin ^2 x}} d x \\
& \left.=\int \frac{\sec ^2 x d x}{\sqrt{1-\tan ^2 x}} \text { \{Multiplying } N^{\prime} r \text { and } D^{\prime} r \text { by } \sec x\right\}
\end{aligned}$
Now putting $\tan x=t \Rightarrow \sec ^2 x d x=d t$, we get the integral $=\sin ^{-1} t=\sin ^{-1}(\tan x)$.
Trick : Since $\frac{d}{d x}\left\{\sin ^{-1}(\tan x)\right\}=\frac{\sec ^2 x}{\sqrt{1-\tan ^2 x}}$
$=\frac{\sec ^2 x \cdot \cos x}{\sqrt{\cos ^2 x-\sin ^2 x}}=\frac{\sec x}{\sqrt{\cos 2 x}} .$
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