Definite Integration Question 23
Question: If $ f(x) $ is an odd function of $ x, $ then $ \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}{f(\cos x)dx} $ is equal to
[MP PET 1998]
Options:
A) 0
B) $ \int_0^{\frac{\pi }{2}}{f(\cos x)dx} $
C) $ 2\int_0^{\frac{\pi }{2}}{f(\sin x)dx} $
D) $ \int_0^{\pi }{f(\cos x)dx} $
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Answer:
Correct Answer: C
Solution:
$ f(\cos x) $ is an even function. $ \because f(\cos (-x))=f(\cos x) $
$ \therefore \int _{-\pi /2}^{\pi /2}{f(\cos x)dx=2\int_0^{\pi /2}{f(\cos x)dx}} $
$ =2\int_0^{\pi /2}{f(\sin x)dx} $ .
BETA
