A) \[\frac{{{\pi }^{2}}}{4}\]
B) \[{{\pi }^{2}}\]
C) zero
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
\[\int_{-a}^{a}{f(x)}\,dx=\left\{ \begin{matrix} {{_{0}^{2}}^{\int_{0}^{a}{f(x)dx,\,\,\text{if}}}} & f(-x)=f(x) \\ 0 & ,\text{if}\,f(-x)=-f(x) \\ \end{matrix} \right.\] Let \[l=\int_{-\pi }^{\pi }{\frac{2x(1+\sin x)}{1+{{\cos }^{2}}x}}\] \[=\int_{-\pi }^{\pi }{\frac{2x}{1+{{\cos }^{2}}x}dx+\int_{-\pi }^{\pi }{\frac{2x\sin x}{1+{{\cos }^{2}}x}dx}}\] \[=0+4\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\] \[\left( \because \frac{2x}{1+{{\cos }^{2}}x}\text{is an odd function} \right)\] \[\therefore l=4\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\] ... (i) \[\Rightarrow \,l=4\int_{0}^{\pi }{\frac{(\pi -x)\sin (\pi -x)}{1+{{\cos }^{2}}(\pi -x)}dx}\] \[\Rightarrow \,l=4\,\int_{0}^{\pi }{\frac{\pi \sin x}{1+{{\cos }^{2}}x}dx-4\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}}\] \[\Rightarrow l=4\pi \int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}}dx=l\] [from Eq. (i] \[\Rightarrow l=2\pi \int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}dx-l}\] \[\Rightarrow \,l=2\pi \int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}dx}\] Put \[\cos x=t\Rightarrow -\sin xdx=dt\] \[\therefore l=-2\pi \int_{1}^{-1}{\frac{1}{1+{{t}^{2}}}dt=2\pi [{{\tan }^{-1}}t]_{-1}^{1}}\] \[=2\pi \left[ \frac{\pi }{4}+\frac{\pi }{4} \right]=2\pi .\frac{\pi }{2}={{\pi }^{2}}\]
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