Question

If $I_1={\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx,I_2={\int }_0^{\pi }x{\sin }^{4}xdx$ then $I_1:I_2=$

• A. $3:4$
• B. $1:2$
• C. $4:3$
• D. $2:3$

Answer 1

The correct option is C $4:3$

$I_1={\int }_0^{\pi }\frac{(\pi -x)\sin (\pi -x)}{1+\left(\cos \right(\pi -x){)}^2}dx$
$=\pi {\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx-{\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$
$2I_1={\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx=2\pi {\int }_0^{\pi /2}\frac{\sin x}{1+{\cos }^{2}x}dx$
$\Rightarrow I_1=\pi {\int }_0^{\pi /2}\frac{\sin x}{1+{\cos }^{2}x}dx=\pi {\int }_0^1\frac{dt}{1+t^2}(t=\cos x)$
$=\pi {\tan }^{-1}t{]}_0^1=\frac{{\pi }^2}{4}$
$I_2={\int }_0^{\pi }(\pi -x){\sin }^{4}xdx$
$=\pi {\int }_0^{\pi }{\sin }^{4}xdx-I_2$
$\Rightarrow 2I_2=2\pi {\int }_0^{\pi /2}{\sin }^{4}xdx=2\pi \cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}$
$\Rightarrow I_2=\frac{3}{16}{\pi }^2$ Therefore

$I_1:I_2=\frac{1}{4}:\frac{3}{16}=4:3$