MHT CET · Maths · Indefinite Integration
\(\int \frac{1}{\cos x \sqrt{\cos 2 x}} d x=\)
(where C is a constant of integration.)
- A \(\sin ^{-1}(\tan x)+C\)
- B \(\log \left(\tan x+\sqrt{\tan ^2 x+1}\right)+C\)
- C \(\tan ^{-1} x+C\)
- D \(\log \left(\tan x+\sqrt{\tan ^2 x-1}\right)+C\)
Answer & Solution
Correct Answer
(A) \(\sin ^{-1}(\tan x)+C\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int \frac{d x}{\cos x \sqrt{\cos 2 x}}=\int \frac{d x}{\cos x \sqrt{\frac{1-\tan ^2 x}{1+\tan ^2 x}}} \\ & \int \frac{d x}{\cos x \frac{\sqrt{1-\tan ^2 x}}{\sec x}}=\int \frac{\sec ^2 d x}{\sqrt{1-\tan ^2 x}} \\ & \int \frac{d t}{\sqrt{1-t^2}}=\sin ^{-1}(t)+C=\sin ^{-1}(\tan x)+C\end{aligned}\)
[let \(\tan \mathrm{x}=\mathrm{t}\) ]
[let \(\tan \mathrm{x}=\mathrm{t}\) ]
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