A) \[\frac{{{\pi }^{2}}}{4}\]
B) \[{{\pi }^{2}}\]
C) \[zero\]
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
Key Idea: \[\int_{-\pi }^{\pi }{f(x)}dx\{=2f(x),\] if \[f(x)\]is an even function. \[=0\], if \[f(x)\]is an odd function. Let \[I=\int_{-\pi }^{\pi }{\frac{2x(1+\sin x)}{1+{{\cos }^{2}}x}dx}\] \[\Rightarrow \] \[I=\int_{-\pi }^{\pi }{\frac{2x}{1+{{\cos }^{2}}x}dx}\] \[+2\int_{-\pi }^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\] \[\Rightarrow \] \[I=0+4\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\] \[\left( \because \,\,\frac{2x}{1+{{\cos }^{2}}x}dx\,\,is\,\,an\,\,odd\,\,function \right)\] \[\Rightarrow \] \[I=4\int_{0}^{\pi }{\frac{(\pi -x)\sin (\pi -x)}{1+{{\cos }^{2}}(\pi -x)}dx}\] \[\Rightarrow \] \[I=4\int_{0}^{\pi }{\frac{(\pi -x)\sin x}{1+{{\cos }^{2}}x}dx}\] \[\Rightarrow \] \[I=4\pi \int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}dx}\] \[-4\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\] \[\Rightarrow \] \[2I=4\pi \int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}dx}\] Put \[\cos x=t\Rightarrow -\sin x\,\,dx=dt\] \[\Rightarrow \] \[\sin x\,\,dx=-dt\] \[\therefore \] \[I=2\pi \int_{1}^{-1}{\frac{-dt}{1+{{t}^{2}}}=2\pi }\int_{-1}^{1}{\frac{dt}{1+{{t}^{2}}}}\] \[=2\pi [{{\tan }^{-1}}t]_{-1}^{1}\] \[=2\pi [{{\tan }^{-1}}(1)-{{\tan }^{-1}}(-1)]\] \[=2\pi \left[ \frac{\pi }{4}+\frac{\pi }{4} \right]=2\pi \times \frac{\pi }{2}={{\pi }^{2}}\]
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