The value of ∫ _ (π)/(3) ^ (π)/(2) (2 + 3 x ) x (1 + x) dx is equal...

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

Now rearranging the above integral we get,

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

$⇒ I = {∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\dfrac{2\sin x}{\sin^{2}x\left(1 + \cos x\right)}dx + {∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\dfrac{3}{\left(2\cos^{2}\dfrac{x}{2}\right)}dx$

$⇒ I = \mathop{\mathop{{∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\dfrac{2\sin x}{\sin^{2}x\left(1 + \cos x\right)} d x}_{⏟}}_{{I}_{1}} + \dfrac{3}{2}\mathop{\mathop{{∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\sec^{2}\left(\dfrac{x}{2}\right) d x}_{⏟}}_{{I}_{2}}$

$⇒ I = \mathop{\mathop{{∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\dfrac{2\sin x}{\sin^{2}x\left(1 + \cos x\right)}dx}_{⏟}}_{{I}_{1}} + \dfrac{3}{2} \times 2\mathop{\mathop{{\left[tan\dfrac{x}{2}\right]}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}}_{⏟}}_{{I}_{2}}$

$⇒ I = \mathop{\mathop{{∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\dfrac{2\sin x}{\sin^{2}x\left(1 + \cos x\right)}dx}_{⏟}}_{{I}_{1}} + 3\left(1 - \dfrac{1}{\sqrt{3}}\right)$

Now solving ${I}_{1}$ we get,

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

$⇒ {I}_{1} = 2{∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\dfrac{1}{\sin x\left(1 + \cos x\right)}dx$

$⇒ {I}_{1} = 2{∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\dfrac{1}{\left(\dfrac{2\tan \left(\dfrac{x}{2}\right)}{1 + \tan^{2}\left(\dfrac{x}{2}\right)}\right)\left(1 + \dfrac{1 - \tan \left(\dfrac{x}{2}\right)}{1 + \tan^{2}\left(\dfrac{x}{2}\right)}\right)}dx$

$⇒ {I}_{1} = \dfrac{1}{2}{∫}_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}}\dfrac{\left[1 + \tan^{2}\left(\dfrac{x}{2}\right)\right]\sec^{2}\left(\dfrac{x}{2}\right)}{\tan \left(\dfrac{x}{2}\right)}dx$

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

$⇒ {I}_{1} = {∫}_{\dfrac{1}{\sqrt{3}}}^{1}\dfrac{\left(1 + {v}^{2}\right)dv}{v}$

$⇒ {I}_{1} = {\left[\log_{e}\left(v\right) + \dfrac{{v}^{2}}{2}\right]}_{\dfrac{1}{\sqrt{3}}}^{1}$

$⇒ {I}_{1} = \log_{e}\left(1\right) - \log_{e}\left(\dfrac{1}{\sqrt{3}}\right) + \dfrac{1}{2}\left(1 - \dfrac{1}{3}\right)$

$⇒ {I}_{1} = \log_{e}\left(\dfrac{1}{\sqrt{3}}\right) + \dfrac{1}{3}$

So,

$I = \dfrac{1}{3} + \log_{e}\sqrt{3} + 3\left(1 - \dfrac{1}{\sqrt{3}}\right)$

$⇒ I = \dfrac{10}{3} - \sqrt{3} + \log_{e}\sqrt{3}$