JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f(x)=\int_0^x g(t) \log _e\left(\frac{1-t}{1+t}\right) d t\), where \(g\) is a continuous odd function. If \(\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right) d x=\left(\frac{\pi}{\alpha}\right)^2-\alpha\), then \(\alpha\) is equal to ..............
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(f(x)=\int_0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t\) \(f(-x)=\int_0^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t\) \(f(-x)=-\int_0^x g(-y) \ln \left(\frac{1+y}{1-y}\right) d y\) \(=-\int_0^x g(y) \ln \left(\frac{1-y}{1+y}\right) d y\) (g is odd)…
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