TS EAMCET · Maths · Indefinite Integration
\(\int \frac{\sqrt{\cos 2 x}}{\sin x} d x=\)
- A \(\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-\tan ^2 x}}{\sqrt{2}-\sqrt{1-\tan ^2 x}}\right|\) \[ -\frac{1}{2} \log \left|\frac{1-\sqrt{1-\tan ^2 x}}{1+\sqrt{1-\tan ^2 x}}\right|+c \]
- B \(\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-\tan ^2 x}}{\sqrt{2}-\sqrt{1-\tan ^2 x}}\right|\) \[ -\frac{1}{2} \log \left|\frac{1+\sqrt{1-\tan ^2 x}}{1-\sqrt{1-\tan ^2 x}}\right|+c \]
- C \(\frac{1}{4 \sqrt{2}} \log \left|\frac{\sqrt{2}-\sqrt{1-\tan ^2 x}}{\sqrt{2}+\sqrt{1-\tan ^2 x}}\right|\) \[ +\frac{1}{2} \log \left|\frac{1-\sqrt{1-\tan ^2 x}}{1+\sqrt{1-\tan ^2 x}}\right|+c \]
- D \(\frac{1}{4 \sqrt{2}} \log \left|\frac{2-\sqrt{1-\tan ^2 x}}{2+\sqrt{1-\tan ^2 x}}\right|\) \[ +\frac{1}{2 \sqrt{2}} \log \left|\frac{1-\sqrt{1-\tan ^2 x}}{1+\sqrt{1-\tan ^2 x}}\right|+c \]
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-\tan ^2 x}}{\sqrt{2}-\sqrt{1-\tan ^2 x}}\right|\) \[ -\frac{1}{2} \log \left|\frac{1+\sqrt{1-\tan ^2 x}}{1-\sqrt{1-\tan ^2 x}}\right|+c \]
Step-by-step Solution
Detailed explanation
We have, \(I=\int \frac{\sqrt{\cos 2 x}}{\sin x} d x\) Put \(\cos 2 x=\frac{1-\tan ^2 x}{1+\tan ^2 x}\) \(\because \quad I=\int \frac{\sqrt{1-\tan ^2 x}}{\sec x \sin x} d x\) Put \(\quad 1-\tan ^2 x=t^2\)…
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