A) 0
B) \[\int_{0}^{\frac{\pi }{2}}{f(\cos x)\,dx}\]
C) \[2\int_{0}^{\frac{\pi }{2}}{f(\sin x)\,dx}\]
D) \[\int_{0}^{\pi }{f(\cos x)\,dx}\]
Correct Answer: C
Solution :
\[f(\cos x)\]is an even function. \[\because f(\cos (-x))=f(\cos x)\] \[\therefore \,\,\int_{-\pi /2}^{\pi /2}{f(\cos x)dx=2\int_{0}^{\pi /2}{f(\cos x)dx}}\]\[=2\int_{0}^{\pi /2}{f(\sin x)dx}\].
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