Definite Integration Question 429
Question: The value of $ \int_0^{\pi /2}{\frac{\sin x}{1+{{\cos }^{2}}x}dx} $ is
[RPET 1995]
Options:
A) $ \pi /2 $
B) $ \pi /4 $
C) $ \pi /3 $
D) $ \pi /6 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_0^{\pi /2}{\frac{\sin x}{1+{{\cos }^{2}}x}}dx $
Put $ \cos x=t $
therefore $ -\sin xdx=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}}} $ .
BETA
