MCQ
$\int_0^a {{x^2}\sin {x^3}\,dx} =$
- A$(1 - \cos {a^3})$
- B$3(1 - \cos {a^3})$
- C$ - \frac{1}{3}(1 - \cos {a^3})$
- ✓$\frac{1}{3}(1 - \cos {a^3})$
$\Rightarrow {x^2}dx = \frac{{dt}}{3}$
$\therefore \,\,\,I = \frac{1}{3}\int_0^{{a^3}} {\sin t\,dt} $
$= - \frac{1}{3}[\cos t]_0^{{a^3}} = - \frac{1}{3}[\cos {a^3} - 1]$
$ = \frac{1}{3}[1 - \cos {a^3}]$.
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