A) \[-\pi \]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) None of these
Correct Answer: C
Solution :
Let\[I=\int_{0}^{\pi }{\frac{xdx}{1+\sin x}}\] ?(i) \[\Rightarrow \] \[I=\int_{0}^{\pi }{\frac{(\pi -x)dx}{1+\sin (\pi -x)}}\] ?(ii) On adding Eqs. (i) and (ii), \[\Rightarrow \] \[2I=\int_{0}^{\pi }{\frac{\pi }{1+\sin x}}dx\] \[=\pi \int_{0}^{\pi }{\frac{1-\sin x}{1+{{\sin }^{2}}x}}dx\] \[=\pi \int_{0}^{\pi }{\frac{1-\sin x}{{{\cos }^{2}}x}}dx\] \[=\pi \int_{0}^{\pi }{({{\sec }^{2}}x-\tan x\sec x)}dx\] \[=\pi [\tan x-\sec x]_{0}^{\pi }\] \[=\pi [0-(-1)-(0-1)]\] \[\Rightarrow \] \[2I=2\pi \] \[\Rightarrow \] \[I=\pi \] \[\Rightarrow \] \[\int_{0}^{\pi }{\frac{xdx}{1+\sin x}}=\pi \]
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