MCQ
$\int_0^{\pi /6} {(2 + 3{x^2})\cos 3x\,dx = } $
- A$\frac{1}{{36}}(\pi + 16)$
- B$\frac{1}{{36}}(\pi - 16)$
- C$\frac{1}{{36}}({\pi ^2} - 16)$
- ✓$\frac{1}{{36}}({\pi ^2} + 16)$
$ = \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)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.