A) \[\frac{{{\pi }^{2}}}{\sqrt{2}}\]
B) \[\frac{{{\pi }^{2}}}{2\sqrt{2}}\]
C) \[2\sqrt{2}\,\pi \]
D) \[\frac{\pi }{2\sqrt{2}}\]
Correct Answer: B
Solution :
\[I=\int\limits_{0}^{\pi }{\frac{x}{1+{{\cos }^{2}}x}}\,dx\,\,=\,\,\int\limits_{0}^{\pi }{\frac{\pi -x}{1+{{\cos }^{2}}(\pi -x)}\,dx}\] \[=\,\,\int\limits_{0}^{\pi }{\frac{\pi -x}{1+{{\cos }^{2}}x}}\] \[=\,\,\int\limits_{0}^{\pi }{\frac{\pi }{1+{{\cos }^{2}}x}}\,dx\,-\,\,\int\limits_{0}^{\pi }{\frac{x}{1+{{\cos }^{2}}\,x}\,\,dx}=\pi \int\limits_{0}^{\pi }{\frac{dx}{1+{{\cos }^{2}}x}-I}\] \[2I=\pi \int\limits_{0}^{\pi }{\frac{1}{1+{{\cos }^{2}}x}}\,dx=2\pi \,\int\limits_{0}^{\frac{\pi }{2}}{\frac{1}{1+{{\cos }^{2}}x}\,dx}\] \[=\,\,2\pi \,\int\limits_{0}^{\frac{\pi }{2}}{\frac{{{\sec }^{2}}x\,\,dx}{2+{{\tan }^{2}}x}}\] Put \[\tan x = t \Rightarrow \,\,se{{c}^{2}}x\,\,dx = dt\] \[2\,I=2\pi \int\limits_{0}^{\infty }{\frac{dt}{2+{{t}^{2}}}}=2\pi .\frac{1}{\sqrt{2}}\left( {{\tan }^{-1}}\frac{t}{\sqrt{2}} \right)_{0}^{\infty }\] \[=\,\,\sqrt{2}\pi \left[ {{\tan }^{-1}}\infty -{{\tan }^{-1}}\,0 \right]\,\,=\sqrt{2}\,\pi \left[ \frac{\pi }{2}-0 \right]=\frac{{{\pi }^{2}}}{\sqrt{2}}\] \[\therefore \,\,I=\frac{{{\pi }^{2}}}{2\sqrt{2}}\]
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