Let f(x) = ∫ g(t)loge(1−t / 1+t)dt ; t ∈ [0, x], ​​where g is a continuous odd function.

Question

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 _____.

Answer 1

Correct answer: 2

\(f(x)=\int_{0}^{x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t\)

\(f(-x)=\int_{0}^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t\)

\(f(-x)=-\int_{0}^{x} g(-y) \ln \left(\frac{1+y}{1-y}\right) d y\)

\(=-\int_{0}^{x} g(y) \ln \left(\frac{1-y}{1+y}\right) d y\)   (g is odd)

\({f}(-{x})=-{f}({x}) \Rightarrow {f}\) is also odd

Now,

\(I=\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^{2} \cos x}{1+e^{x}}\right) d x\quad ....(1)\) 

\(I=\int_{-\pi / 2}^{\pi / 2}\left(f(-x)+\frac{x^{2} e^{x} \cos x}{1+e^{x}}\right) d x\quad ....(2)\)

\(2 I=\int_{-\pi / 2}^{\pi / 2} x^{2} \cos x d x=2 \int_{0}^{\pi / 2} x^{2} \cos x d x\)

\( I=\left(x^{2} \sin x\right)_{0}^{\pi / 2}-\int_{0}^{\pi / 2} 2 x \sin x d x\)

\( =\frac{\pi^{2}}{4}-2\left(-x \cos x+\int \cos x d x\right)_{0}^{\pi / 2}\)

\(=\frac{\pi^{2}}{4}-2(0+1)\)

\(=\frac{\pi^{2}}{4}-2 \)

\(\Rightarrow\left(\frac{\pi}{2}\right)^{2}-2\)

\( \therefore \alpha=2\)