Question

If $\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{1+{\sin }^{2}x}dx=\frac{\pi }{a}\ln \left(\frac{b+1}{c+1}\right);a,b,c\in R.$ Then which of the following is/are correct ?

• A. $a+bc=2$
• B. $4b+ac=4$
• C. $a-b-c=2$
• D. $a+b+c=2$

Answer 1

Answer: (A), (B), (C)

$a-b-c=2$

Let $I=\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{1+{\sin }^{2}x}dx\cdots \left(i\right)$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow I=\underset{0}{\overset{\pi }{\int }}\frac{(\pi -x)\sin x}{1+{\sin }^{2}x}dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right):$
$\Rightarrow 2I=\underset{0}{\overset{\pi }{\int }}\frac{\pi \sin x}{1+{\sin }^{2}x}dx\Rightarrow I=-\frac{\pi }{2}\underset{0}{\overset{\pi }{\int }}\frac{-\sin x}{2-{\cos }^{2}x}dx$
Substitute $\cos x=t\Rightarrow -\sin xdx=dt$
$\Rightarrow I=-\frac{\pi }{2}\underset{1}{\overset{-1}{\int }}\frac{dt}{2-t^2}\Rightarrow I=\frac{\pi }{4\sqrt{2}}\ln {\left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|}_{-1}^1\Rightarrow I=\frac{\pi }{2\sqrt{2}}\ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$
So, $a=2\sqrt{2},b=\sqrt{2},c=\sqrt{2}-2$