A) \[\frac{2}{3}\]
B) \[\frac{\pi }{4}\]
C) 1
D) \[\frac{3\pi }{4}\]
Correct Answer: D
Solution :
\[\int\limits_{0}^{\pi /2}{\left( \frac{1+\sin 3x}{1+2\sin x} \right)\,dx=\int\limits_{0}^{\pi /2}{\left( \frac{1+3\sin x\,-4{{\sin }^{3}}x}{1+2\sin x} \right)}}\] = 1You need to login to perform this action.
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