MCQ
$\int_{}^{} {{{\sin }^2}x\cos x\;dx} $=
- A$\frac{{{{\cos }^2}x}}{2} + c$
- B$\frac{{{{\sin }^2}x}}{3} + c$
- ✓$\frac{{{{\sin }^3}x}}{3} + c$
- D$ - \frac{{{{\cos }^2}x}}{2} + c$
Put $\sin x = t \Rightarrow \cos x\,dx = dt$
$\therefore \,\,\,I = \int_{}^{} {{t^2}dt} = \frac{{{t^3}}}{3} + c = \frac{{{{\sin }^3}x}}{3} + c$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.