AP EAMCET · Maths · Indefinite Integration
There exist \(\theta\) such that \(\mathrm{a}>|\sec \theta|\), then \(\int \frac{d x}{1+a \cos x}=\)
- A \(\frac{1}{\sqrt{a^2-1}} \tan ^{-1}\left(\frac{\sqrt{a-1}}{\sqrt{a+1}} \tan \frac{x}{2}\right)+C\)
- B \(\frac{1}{\sqrt{a^2-1}} \tan ^{-1}\left(\frac{\sqrt{1-a}}{\sqrt{1+a}} \tan \frac{x}{2}\right)+C\)
- C \(\frac{1}{\sqrt{a^2-1}} \log \left(\frac{\sqrt{a+1} \cos \frac{x}{2}-\sqrt{a-1} \sin \frac{x}{2}}{\sqrt{a-1} \cos \frac{x}{2}+\sqrt{a-1} \sin \frac{x}{2}}\right)+C\)
- D \(\frac{1}{\sqrt{a^2-1}} \log \left(\frac{\sqrt{a+1} \cos \frac{x}{2}+\sqrt{a-1} \sin \frac{x}{2}}{\sqrt{a+1} \cos \frac{x}{2}-\sqrt{a-1} \sin \frac{x}{2}}\right)+C\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{\sqrt{a^2-1}} \log \left(\frac{\sqrt{a+1} \cos \frac{x}{2}+\sqrt{a-1} \sin \frac{x}{2}}{\sqrt{a+1} \cos \frac{x}{2}-\sqrt{a-1} \sin \frac{x}{2}}\right)+C\)
Step-by-step Solution
Detailed explanation
Given, \(\int \frac{1}{1+a \cos (x)} d x\) We are substituting \(\cos (x)=\frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)}\).…
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