A) 0
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
Correct Answer: A
Solution :
Given that, \[I=\int_{0}^{\pi /2}{\frac{\cos x-\sin x}{1+\cos x\sin x}dx}\] ?. (i) Putting \[x=\left( \frac{\pi }{2}-x \right)\] in equation (i), we get \[I=\int_{0}^{\pi /2}{\frac{\cos (\pi /2-x)-\sin \,(\pi /2-x)}{1+\cos \,(\pi /2-x)\sin \,(\pi /2-x)}}dx\] \[\Rightarrow \] \[I=\int_{0}^{\pi /2}{-\left( \frac{\cos x-\sin x}{1+\cos x\sin x} \right)dx}\] ?. (ii) On adding equation (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{\left( \frac{\cos x-\sin x}{1+\cos x\sin x}-\frac{\cos x-\sin x}{1+\cos x\sin x} \right)dx}\] \[=\int_{0}^{\pi /2}{0\,.\,\,dx=0}\] \[\Rightarrow \] \[I=0\]
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