Question

If $I_1={\int }_0^{\pi }\frac{x\text{sin}x}{1+{\text{cos}}^{2}x}dx\text{, }{I}_2={\int }_0^{\pi }x{\text{sin}}^{4}xdx$ then, $I_1\text{:}{I}_2$ is equal to

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

Answer 1

Answer: (C)

Given, $I_1={\int }_0^{\pi }\frac{\left(\pi -x\right)\text{sin}\left(\pi -x\right)}{1+{\left(\text{cos}\left(\pi -x\right)\right)}^2}dx\left({\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right)$
$=\pi {\int }_0^{\pi }\frac{sinx}{1+{\text{cos}}^{2}x}dx-{\int }_0^{\pi }\frac{xsinx}{1+{\text{cos}}^{2}x}dx$
$2I_1=\pi {\int }_0^{\pi }\frac{sinx}{1+{\text{cos}}^{2}x}dx=2\pi {\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+{\text{cos}}^{2}x}dx$
$\Rightarrow I_1=\pi {\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+{\left(\text{cos}\right)}^{2}x}dx=\pi {\int }_0^1\frac{dt}{1+t^2}\left(t=cosx\right)$
$={\left[\pi {\text{tan}}^{-1}t\right]}_0^1=\frac{{\pi }^2}{4}$
$I_2={\int }_0^{\pi }\left(\pi -x\right){\left(\text{sin}\right)}^{4}xdx$
$=\pi {\int }_0^{\pi }{\text{sin}}^{4}xdx-I_2$
$\Rightarrow 2I_2=2\pi {\int }_0^{\pi /2}{\text{sin}}^{4}xdx=2\pi \cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}$
$\Rightarrow I_2=\frac{3}{16}{\pi }^2$
Therefore, $I_1:I{}_2=\frac{1}{4}:\frac{3}{16}=4:3$