A) \[\frac{{{\pi }^{2}}}{4}\]
B) \[{{\pi }^{2}}\]
C) zero
D) \[\frac{\pi }{2}\]
Correct Answer: A
Solution :
\[I=\int_{-\pi }^{\pi }{\frac{2x(1+\sin x)}{1+{{\cos }^{2}}x}}dx\] \[I=\int_{-\pi }^{\pi }{\frac{2x}{1+{{\cos }^{2}}x}}dx+2\int_{-\pi }^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}}dx\] \[I=0+4\int_{0}^{\pi }{\frac{x\sin xdx}{1+{{\cos }^{2}}x}}\] \[(\because \int_{-\pi }^{\pi }{\frac{2x}{1+{{\cos }^{2}}x}}dx\]is an odd function) \[I=4\int_{0}^{\pi }{\frac{(\pi -x)\sin (\pi -x)}{1+{{\cos }^{2}}(\pi -x)}}dx\] \[I=4\int_{0}^{\pi }{\frac{(\pi -x)\sin x}{1+{{\cos }^{2}}(\pi -x)}}dx\] \[I=4\int_{0}^{\pi }{\frac{(\pi -x)\sin x}{1+{{\cos }^{2}}x}}dx\] \[\Rightarrow \]\[I=4\pi \int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}}dx-4\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}}dx\] \[2I=4\pi \int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}}dx\] put \[cos\text{ }x=t\] \[-\sin x\,dx=dt\]\[\Rightarrow \]\[\sin xdx=-dt\] \[I=2\pi \int_{1}^{1}{\frac{-dt}{1+{{t}^{2}}}}=2\pi \int_{-1}^{1}{\frac{dt}{1+{{t}^{2}}}}\] \[=2\pi \left[ {{\tan }^{-1}} \right]_{-1}^{1}\] \[=2\pi [{{\tan }^{-1}}(1)-{{\tan }^{-1}}(-1)]\] \[=2\pi \left[ \frac{\pi }{4}+\frac{\pi }{4} \right]=2\times \frac{\pi }{2}={{\pi }^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
