A) 1
B) 2
C) -1
D) -2
Correct Answer: B
Solution :
Let\[I=\int_{0}^{\pi }{\frac{1}{1+\sin x}dx}\] \[=\int_{0}^{\pi }{\frac{1}{1+\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}dx}\] \[=\int_{0}^{\pi }{\frac{{{\sec }^{2}}\frac{x}{2}}{\left( 1+\tan \frac{x}{2} \right)}dx}\] Put\[\tan \frac{x}{2}=t\] \[\Rightarrow \] \[\frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx=dt\] \[\therefore I=\int_{0}^{\infty }{\frac{2\,\,dt}{{{(1+t)}^{2}}}={{\left[ -\frac{2}{1+t} \right]}_{0}}=2}\]
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