Let f(x)= _0^x g(t) _e (1-t 1+t ) d t, where g is a continuous od… - Vidyadip

MCQ

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
$1$
✓
$2$
D
$3$

✓

Answer

Correct option: C.

$2$

c
$f(x)=\int_0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t$

$f(-x)=\int_0^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t$

$f(-x)=-\int_0^x g(-y) \ln \left(\frac{1+y}{1-y}\right) d y$

$=-\int_0^x g(y) \ln \left(\frac{1-y}{1+y}\right) d y$ (g is odd)

$f(-x)=-f(x) \Rightarrow f$ is also odd

Now,

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

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

$2 I=\int_{-\pi / 2}^{\pi / 2} x^2 \cos x d x=2 \int_0^{\pi / 2} x^2 \cos x d x$

$ 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$

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