Mathematics - Definite Integration Question with Solution | TestHub

Let, $I={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{2+3\sin x}{\sin x\left(1+\cos x\right)}dx$

Now rearranging the above integral we get,

$I={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{2}{\sin x\left(1+\cos x\right)}dx+{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{3}{\left(1+\cos x\right)}dx$

$\Rightarrow I={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{2\sin x}{{\sin }^{2}x\left(1+\cos x\right)}dx+{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{3}{\left(2{\cos }^2{\displaystyle \frac{x}{2}}\right)}dx$

$\Rightarrow I=\underset{I_1}{\underbrace{{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{2\sin x}{{\sin }^{2}x\left(1+\cos x\right)}dx}}+\frac{3}{2}\underset{I_2}{\underbrace{{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}{\sec }^2\left(\frac{x}{2}\right)dx}}$

$\Rightarrow I=\underset{I_1}{\underbrace{{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{2\sin x}{{\sin }^{2}x\left(1+\cos x\right)}dx}}+\frac{3}{2}\times 2\underset{I_2}{\underbrace{{\left[\tan \frac{x}{2}\right]}_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}}}$

$\Rightarrow I=\underset{I_1}{\underbrace{{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{2\sin x}{{\sin }^{2}x\left(1+\cos x\right)}dx}}+3\left(1-\frac{1}{\sqrt{3}}\right)$

Now solving $I_1$ we get,

$I_1={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{2\sin x}{{\sin }^{2}x\left(1+\cos x\right)}dx$

$\Rightarrow I_1=2{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{1}{\sin x\left(1+\cos x\right)}dx$

$\Rightarrow I_1=2{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{1}{\left({\displaystyle \frac{2\tan \left({\displaystyle \frac{x}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{x}{2}}\right)}}\right)\left(1+\frac{1-\tan \left({\displaystyle \frac{x}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{x}{2}}\right)}\right)}dx$

$\Rightarrow I_1=\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{\left[1+{\tan }^2\left(\frac{{\displaystyle x}}{{\displaystyle 2}}\right)\right]{\sec }^2\left({\displaystyle \frac{x}{2}}\right)}{{\displaystyle \tan \left(\frac{{\displaystyle x}}{{\displaystyle 2}}\right)}}dx$

Put $\tan \left(\frac{x}{2}\right)=v\Rightarrow \frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx=dv$

$\Rightarrow I_1={\int }_{\frac{1}{\sqrt{3}}}^1\frac{\left(1+v^2\right)dv}{v}$

$\Rightarrow I_1={\left[{\log }_e\left(v\right)+\frac{v^2}{2}\right]}_{\frac{1}{\sqrt{3}}}^1$

$\Rightarrow I_1={\log }_e\left(1\right)-{\log }_e\left(\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(1-\frac{1}{3}\right)$

$\Rightarrow I_1={\log }_e\left(\frac{1}{\sqrt{3}}\right)+\frac{1}{3}$

So,

$I=\frac{1}{3}+{\log }_e\sqrt{3}+3\left(1-\frac{1}{\sqrt{3}}\right)$

$\Rightarrow I=\frac{10}{3}-\sqrt{3}+{\log }_e\sqrt{3}$