JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int \frac{1}{ x } \sqrt{\frac{1- x }{1+ x }} dx = g ( x )+ c , g (1)=0\), then \(g \left(\frac{1}{2}\right)\) is equal to
- A \(\log _{e}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi}{3}\)
- B \(\log _{e}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)+\frac{\pi}{3}\)
- C \(\log _{ e }\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)-\frac{\pi}{3}\)
- D \(\frac{1}{2} \log _{ e }\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)-\frac{\pi}{6}\)
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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