A) \[\pi /4\]
B) \[\pi /2\]
C) 0
D) \[\pi \]
Correct Answer: A
Solution :
| [a] : Let \[I=\int\limits_{0}^{\pi /2}{\frac{dx}{1+\cot x}}=\int\limits_{0}^{\pi /2}{\frac{\cos xdx}{\cos x+\sin x}}\] |
| Replacing x by \[\frac{\pi }{2}-x\], we get |
| \[I=\int\limits_{0}^{\pi /2}{\frac{\cos \left( \frac{\pi }{2}-x \right)dx}{\cos \left( \frac{\pi }{2}-x \right)+\sin \left( \frac{\pi }{2}-x \right)}}\] |
| \[I=\int\limits_{0}^{\pi /2}{\frac{\sin x\,dx}{\sin x+\cos x}}\] |
| \[\therefore \]\[I+I=\int\limits_{0}^{\pi /2}{\frac{\cos xdx}{\cos x+\sin x}}+\int\limits_{0}^{\pi /2}{\frac{\sin xdx}{\sin x+\cos x}}\] |
| \[=\int\limits_{0}^{\pi /2}{dx}=[x]_{0}^{\pi /2}=\frac{\pi }{2}\] |
| So,\[2I=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}\Rightarrow \int\limits_{0}^{\pi /2}{\frac{dx}{1+\cot x}=\frac{\pi }{4}}\] |
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