Definite Integration Question 375
Question: $ \int_0^{\pi /6}{(2+3x^{2})\cos 3xdx=} $
[DSSE 1985]
Options:
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) $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ I=\int_0^{\pi /6}{( 2+3x^{2} )\cos 3xdx} $
$ =[ \frac{\sin 3x}{3}(2+3x^{2}) ]_0^{\pi /6}-\int_0^{\pi /6}{\frac{\sin 3x}{3}}.6x.dx $
$ =\frac{1}{36}({{\pi }^{2}}+16) $ .
BETA
