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$\frac{1}{{36}}({\pi ^2} - 16)$
- ✓$\frac{1}{{36}}({\pi ^2} + 16)$
✓
Answer
Correct option: D.
$\frac{1}{{36}}({\pi ^2} + 16)$
d
(d) Let $I = \int_0^{\pi /6} {\left( {2 + 3{x^2}} \right)\cos 3x\,dx} $
(d) 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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