A) \[1+\pi (2-\pi )\]
B) \[1-\pi (\pi -2)\]
C) \[1+\pi (2+\pi )\]
D) \[1-\pi (2+\pi )\]
Correct Answer: A
Solution :
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 \[I=1+\pi (-\pi +2)\]
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