A) \[\frac{1}{36}(\pi +16)\]
B) \[\frac{1}{36}(\pi -16)\]
C) \[\frac{1}{36}({{\pi }^{2}}-16)\]
D) \[\frac{1}{36}({{\pi }^{2}}+16)\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi /6}{\left( 2+3{{x}^{2}} \right)\cos 3x\,dx}\] \[=\left[ \frac{\sin 3x}{3}(2+3{{x}^{2}}) \right]_{0}^{\pi /6}-\int_{0}^{\pi /6}{\frac{\sin 3x}{3}}.6x.dx\] \[=\frac{1}{36}({{\pi }^{2}}+16)\].
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