Matrix Matching : Integrals and Values 

Option (A) $\underset{0}{\overset{\pi }{\int }}\frac{1}{4-{\sin }^{2}x}.dx\Rightarrow I=\underset{0}{\overset{\pi }{\int }}\frac{1}{4-{\sin }^{2}x}.dx=\underset{0}{\overset{\pi }{\int }}\frac{1}{4-\left(\frac{1-\cos 2x}{2}\right)}.dx$

$\underset{0}{\overset{\pi }{\int }}\frac{1}{\frac{8-1+\cos 2x}{2}}.dx=\underset{0}{\overset{\pi }{\int }}\frac{2}{7+\cos 2x}.dx$

$=\underset{0}{\overset{\pi }{\int }}\frac{2}{7+\left(\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\right)}.dx=\underset{0}{\overset{\pi }{\int }}\frac{2[1+{\tan }^{2}x]}{8+6{\tan }^{2}x}dx$

$=\underset{0}{\overset{\pi }{\int }}\frac{1+{\tan }^{2}x}{4+3{\tan }^{2}x}.dx=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1+{\tan }^{2}x}{4+3{\tan }^{2}x}dx=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sec }^{2}x}{4+3{\tan }^{2}x}.dx$

Let $\tan x=tx=0\Rightarrow t=0$

${\sec }^{2}xdx=dt$ $x=\frac{\pi }{2}\Rightarrow t=\infty$

$=2\underset{0}{\overset{\infty }{\int }}\frac{dt}{4+3t^2}=2\times \frac{1}{3}\underset{0}{\overset{\infty }{\int }}\frac{dt}{{\left(\frac{2}{\sqrt{3}}\right)}^2+t^2}$

$=\frac{2}{3}\times \frac{1}{\frac{2}{\sqrt{3}}}\times {\left({\text{Tan}}^{-1}\left(\frac{t}{\frac{2}{\sqrt{3}}}\right)\right)}_0^{\infty }$ 

$=\frac{1}{\sqrt{3}}\times (\frac{\pi }{2}-0)=\frac{\pi }{2\sqrt{3}}$

Option (B) : $I=\underset{0}{\overset{\pi }{\int }}\frac{x}{4-{\sin }^{2}x}.dx=\underset{0}{\overset{\pi }{\int }}\frac{(\pi -x)}{4-{\sin }^{2}x}.dx=\underset{0}{\overset{\pi }{\int }}\frac{\pi }{4-{\sin }^{2}x}.dx-\underset{0}{\overset{\pi }{\int }}\frac{x}{4-{\sin }^{2}x}.dx$

$=\underset{0}{\overset{\pi }{\int }}\frac{\pi }{4-{\sin }^{2}x}.dx-I$

$2I=\pi \left(\frac{\pi }{2\sqrt{3}}\right)\Rightarrow I=\frac{{\pi }^2}{4\sqrt{3}}$

Option  ( C ) : $\underset{0}{\overset{\pi }{\int }}\frac{\sin x}{4-{\sin }^{2}x}.dx$

$I=\underset{0}{\overset{\pi }{\int }}\frac{\sin x}{4-{\sin }^{2}x}.dx=\underset{0}{\overset{\pi }{\int }}\frac{\sin x}{3+{\cos }^{2}x}dx$

$=\frac{-1}{\left(\sqrt{3}\right)}{\text{\hspace{0.17em}\hspace{0.17em}Tan}}^{-1}{\left(\frac{t}{\sqrt{3}}\right)}^{-1}$

$=\frac{-1}{\sqrt{3}}({\text{Tan}}^{-1}\left(\frac{-1}{\sqrt{3}}\right)-{\text{Tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right))$

$=\frac{-1}{\sqrt{3}}(\frac{-\pi }{6}-\frac{\pi }{6})=\frac{\pi }{3\sqrt{3}}=\frac{\pi }{3\sqrt{3}}$

$\underset{0}{\overset{\pi }{\int }}\frac{4\sin x}{4-{\sin }^{2}x}.dx=\frac{\pi }{3\sqrt{3}}$

Option (D) :

$I=\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{4-{\sin }^{2}x}=\underset{0}{\overset{\pi }{\int }}\frac{(\pi -x)\sin (\pi -x)}{4-{\sin }^2}=\underset{0}{\overset{\pi }{\int }}\frac{\pi \sin x}{4-{\sin }^{2}x}-\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{4-{\sin }^{2}x}$

$I=\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin x}{4-{\sin }^{2}x}-I\Rightarrow 2I=\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin x}{4-{\sin }^{2}x}$

$\Rightarrow 2I=\frac{\pi \times \pi }{3\sqrt{3}}\Rightarrow I=\frac{{\pi }^2}{6\sqrt{3}}$