, then the value of \[\int\limits_{0}^{\pi /2}{f(x)dx}\] is
A) 3
B) 2/3
C) 1/3
D) 0
Correct Answer: C
Solution :
[c] : Applying \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}}-{{C}_{3}}\], we get 
\[=3\sin x-4{{\sin }^{3}}x=\sin 3x\] Now,\[\int\limits_{0}^{\pi /2}{\sin 3xdx}=\left[ \frac{-\cos 3x}{3} \right]_{0}^{\pi /2}=\left[ \frac{-0+1}{3} \right]=\frac{1}{3}\]
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