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) \[I=\int_{0}^{\pi }{\frac{(\pi -x)dx}{1+\sin (\pi -x)}}\] \[I=\int_{0}^{\pi }{\frac{(\pi -x)dx}{1+\sin x}}\] ?.. (ii), \[\left\{ \because \,\int_{0}^{a}{f(x)\,dx}=\int_{0}^{a}{f(a-x)\,dx} \right\}\,\] Adding (i) and (ii), we get \[2I=\int_{0}^{\pi }{\frac{\pi \,dx}{1+\sin x}}\] \[2I=\pi \int_{0}^{\pi }{\frac{1-\sin x}{(1+\sin x)(1-\sin x)}dx}\] \[2I=\pi \int_{0}^{\pi }{\frac{1-\sin x}{{{\cos }^{2}}x}}dx=\pi \int_{0}^{\pi }{({{\sec }^{2}}x-\sec x\tan x)dx}\] \[2I=\pi [\tan x-\sec x]_{0}^{\pi }=\pi [0-(-1)-(0-1)]\],\[2I=2\pi \] \ \[I=\pi \].B
You need to login to perform this action.
You will be redirected in
3 sec
