A) \[\pi /2\]
B) \[\pi /4\]
C) \[\pi /3\]
D) \[\pi /6\]
Correct Answer: B
Solution :
\[I=\int_{0}^{\pi /2}{\frac{\sin x}{1+{{\cos }^{2}}x}}dx\] Put \[\cos x=t\]Þ \[-\sin x\,dx=dt\] Then \[I=\int_{1}^{0}{\frac{-dt}{1+{{t}^{2}}}=\int_{0}^{1}{\frac{dt}{1+{{t}^{2}}}=[{{\tan }^{-1}}t]_{0}^{1}=\frac{\pi }{4}}}\].You need to login to perform this action.
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