A) \[\frac{1}{2}+\frac{1}{\pi +2}-A\]
B) \[\frac{1}{\pi +2}-A\]
C) \[1+\frac{1}{\pi +2}-A\]
D) \[A-\frac{1}{2}-\frac{1}{\pi +2}\]
Correct Answer: A
Solution :
[a] \[I=\int_{0}^{\pi /2}{\frac{\sin 2x}{x+1}dx.}\] Put \[x=y/2.\]Then, \[I=\int_{0}^{\pi }{\frac{\sin y}{y+2}dy}\] \[=\left( \frac{-\cos y}{y+2} \right)_{0}^{\pi }-\int\limits_{0}^{\pi }{\frac{\cos y}{{{(y+2)}^{2}}}dy}\] (Integrating by parts) \[=\frac{1}{\pi +2}+\frac{1}{2}-A\]You need to login to perform this action.
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