Question

${\int }_0^{\frac{\pi }{6}}(2+3x^2)cos3xdx=$

• A. $\frac{1}{36}(\pi +16)$
• B. $\frac{1}{36}(\pi -16)$
• C. $\frac{1}{36}({\pi }^2-16)$
• D. $\frac{1}{36}({\pi }^2+16)$

Answer 1

The correct option is D $\frac{1}{36}({\pi }^2+16)$

$LetI={\int }_0^{\frac{\pi }{6}}(2+3x^2)cos3xdx$
$={\left[\frac{sin3x}{3}\right(2+3x^2\left)\right]}_0^{\frac{\pi }{6}}-{\int }_0^{\frac{\pi }{6}}\frac{sin3x}{3}.6x.dx$
$=\frac{1}{36}({\pi }^2+16)$