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JEE Mains · Maths · STD 12 - 7.1 indefinite integral

यदि \(\int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} d x=g(x)+c, g(1)=01\) है, तो \(g \left(\frac{1}{2}\right)\) का मान होगा :

  1. A \(\log _{e}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi}{3}\)
  2. B \(\log _{e}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)+\frac{\pi}{3}\)
  3. C \(\log _{ e }\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)-\frac{\pi}{3}\)
  4. D \(\frac{1}{2} \log _{ e }\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)-\frac{\pi}{6}\)
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Answer & Solution

Correct Answer

(A) \(\log _{e}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi}{3}\)

Step-by-step Solution

Detailed explanation

\(\int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} d x=g(x)+c\) Put \(x=\cos 2 \theta\) \(d x=-2 \sin 2 \theta \cdot d \theta\) \(=\int \frac{1}{\cos 2 \theta} \tan \theta(-4 \sin \theta \cdot \cos \theta) d \theta\)…
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