MHT CET · Maths · Indefinite Integration
\(\int \frac{d x}{\cos x \sqrt{\cos 2 x}}=\)
- A \(\frac{1}{2} \log \left|\tan \left(\frac{\pi}{4}+x\right)\right|+c\)
- B \(\frac{1}{2} \log \left|\frac{1-\tan x}{1+\tan x}\right|+c\)
- C \(2 \log \left|\frac{1+\tan x}{1-\tan x}\right|+c\)
- D \(\sin ^{-1}(\tan x)+\mathrm{c}\)
Answer & Solution
Correct Answer
(D) \(\sin ^{-1}(\tan x)+\mathrm{c}\)
Step-by-step Solution
Detailed explanation
Let
\(\begin{aligned} I &=\int \frac{d x}{\cos x \sqrt{\cos 2 x}} \\ &=\int \frac{d x}{\cos x \cdot \cos x \sqrt{\frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x}}}=\int \frac{d x}{\cos ^{2} x \sqrt{1-\tan ^{2} x}} \\ I &=\int \frac{\sec ^{2} x}{\sqrt{1-\tan ^{2} x}} d x \end{aligned}\)
Put \(\tan x=t \quad \Rightarrow \sec ^{2} d x=d t\)
\(\therefore I=\int \frac{d t}{\sqrt{1-t^{2}}}=\sin ^{-1}(t)+c=\sin ^{-1}(\tan x)+c\)
\(\begin{aligned} I &=\int \frac{d x}{\cos x \sqrt{\cos 2 x}} \\ &=\int \frac{d x}{\cos x \cdot \cos x \sqrt{\frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x}}}=\int \frac{d x}{\cos ^{2} x \sqrt{1-\tan ^{2} x}} \\ I &=\int \frac{\sec ^{2} x}{\sqrt{1-\tan ^{2} x}} d x \end{aligned}\)
Put \(\tan x=t \quad \Rightarrow \sec ^{2} d x=d t\)
\(\therefore I=\int \frac{d t}{\sqrt{1-t^{2}}}=\sin ^{-1}(t)+c=\sin ^{-1}(\tan x)+c\)
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