A) \[\frac{{{\pi }^{2}}}{4}\]
B) \[\frac{{{\pi }^{2}}}{2}\]
C) \[\frac{3{{\pi }^{2}}}{2}\]
D) \[\frac{{{\pi }^{2}}}{3}\]
Correct Answer: A
Solution :
Let I =\[\int_{0}^{\pi }{\frac{x\tan x}{\sec x+\cos x}dx}=\int_{0}^{\pi }{\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\cos (\pi -x)}dx}\] It gives \[I=\frac{\pi }{2}\int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}}dx\] Now put \[\cos x=t\]and solve, we get \[I=\frac{\pi }{2}\times \frac{\pi }{2}=\frac{{{\pi }^{2}}}{4}\].
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