Q. $\int_{}^{}{\sin^{3}x.\cos^{2}x}\ dx$ is equal to
A) $\frac{\sin^{5}x}{5} - \frac{\sin^{3}x}{3} + c$
B) $\frac{\sin^{5}x}{5} + \frac{\sin^{3}x}{3} + c$
C) $\frac{\cos^{5}x}{5} - \frac{\cos^{3}x}{3} + c$
D) $\frac{\cos^{5}x}{5} + \frac{\cos^{3}x}{3} + c$
Explanation
(c)
Let $I = \int_{}^{}{(1 - \cos^{2}x})\cos^{2}x.\sin x\ dx$
Put $\cos x = t$
$\Rightarrow - \sin x\ dx = dt$
$\therefore\ I = - \int_{}^{}{(1 - t^{2})}t^{2}dt = \int_{}^{}{(t^{4} - t^{2})}dt$
$= \frac{t^{5}}{5} - \frac{t^{3}}{3} + c$
$= \frac{\left( \cos x \right)^{5}\ }{5} - \frac{\left( \cos x \right)^{3}}{3} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.