A) \[\frac{\pi }{2}-1\]
B) \[\pi \left( \frac{\pi }{2}+1 \right)\]
C) \[\frac{\pi }{2}+1\]
D) \[\pi \left( \frac{\pi }{2}-1 \right)\]
Correct Answer: D
Solution :
\[I=\int_{0}^{\pi }{\frac{x\tan x}{\sec x+\tan x}dx=\int_{0}^{\pi }{\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\tan (\pi -x)}}dx}\] Þ \[2I=\frac{\pi }{2}\int_{0}^{\pi }{\frac{\tan x}{\sec x+\tan x}dx=\frac{\pi }{2}\int_{0}^{\pi }{\frac{\sin x}{1+\sin x}dx}}\] =\[\frac{\pi }{2}\left[ \int_{0}^{\pi }{1dx-\int_{0}^{\pi }{\frac{dx}{1+\sin x}}} \right]\] On solving, we get \[I=\frac{{{\pi }^{2}}}{2}-\pi =\pi \left( \frac{\pi }{2}-1 \right)\].
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