Question

If the value of $ {{\int_{0}^{\pi }{\left( \frac{x}{1+\sin \,x} \right)}}^{2}}dx=1, $ then the value of the integral $ \int_{0}^{\pi }{\left[ \frac{2{{x}^{2}}\,{{\cos }^{2}}\,x/2}{{{(1+\sin \,x)}^{2}}} \right]}dx $ is equal to

• A. $ 1+\pi (2-\pi ) $
• B. $ 1-\pi (\pi -2) $
• C. $ 1+\pi (2+\pi ) $
• D. $ 1-\pi (2+\pi ) $

Answer 1

Answer: (A)

Given that, $ \int_{0}^{\pi }{{{\left( \frac{x}{1+\sin x} \right)}^{2}}\,\,dx=1} $ ..(i)
Let $ I=\int_{0}^{\pi }{\left[ \frac{2{{x}^{2}}\,{{\cos }^{2}}\,(x/2)}{{{(1+\sin \,x)}^{2}}} \right]\,\,dx} $
$ \therefore $ $ I=\int_{0}^{\pi }{\frac{{{x}^{2}}(1+\cos \,x)}{{{(1+\sin \,x)}^{2}}}}dx $
$=\int_{0}^{\pi }{{{\left( \frac{x}{1+\sin x} \right)}^{2}}dx+\int_{0}^{\pi }{\frac{{{x}^{2}}\,\cos x}{(1+\sin x)}}\,\,dx} $
$ \Rightarrow $ $ I=1+{{I}_{2}} $ ..(ii)
Where $ I=1+{{I}_{2}} $
$ \Rightarrow $ $ {{I}_{2}}=-\int_{0}^{\pi }{\frac{{{(\pi -x)}^{2}}\,\cos x}{{{(1+\sin x)}^{2}}}}dx $
$ \Rightarrow $ $ 2{{I}_{2}}=\int_{0}^{\pi }{\frac{(-{{\pi }^{2}}+2\pi x)}{{{(1+\sin \,x)}^{2}}}\,.\,dx} $
$ \Rightarrow $ $ 2{{I}_{2}}=-{{\pi }^{2}}\int_{0}^{\pi }{\frac{\cos \,x}{{{(1+\sin \,x)}^{2}}}\,.\,dx} $ $ +2\pi \int_{0}^{\pi }{\frac{x(\cos x)}{{{(1+\sin \,x)}^{2}}}}dx $
$={{\pi }^{2}}\left[ \frac{1}{1+\sin x} \right]_{0}^{\pi }+2\pi \,\int_{0}^{\pi }{\frac{x\,\,\cos \,x}{{{(1+\,\sin x)}^{2}}}}dx $
$ \Rightarrow $ $ {{I}_{2}}=\frac{{{\pi }^{2}}}{2}[1-1]+{{I}_{3}} $ ?..(iii) Where $ {{I}_{3}}=\pi \int_{0}^{\pi }{\frac{x\,\cos \,x}{{{(1+\sin \,x)}^{2}}}\,dx} $
$=\pi \left[ \left\{ -\frac{x}{1+\sin \,x)} \right\}_{0}^{\pi }+\int_{0}^{\pi }{\frac{1}{1+\sin \,x}}\,\,dx \right] $
$=\pi \left[ -\pi +\int_{0}^{\pi }{\frac{{{\sec }^{2}}\,x/2}{{{(1+\tan \,x/2)}^{2}}}dx} \right] $
$=\pi \left[ -\pi -2\left( \frac{1}{1+\tan \frac{x}{2}} \right)_{0}^{\pi } \right] $
$=\pi (-\pi +2) $ On putting this value in Eq. (iii), we get $ {{I}_{2}}=\pi (-\pi /2) $ On putting this value in Eq. (ii), we get $ I=1+\pi (-\pi +2) $