Question

Solve :- $\int {\sin }^{3}xdx$

Answer 1

$\int {\sin }^{3}xdx$
$=\int {\sin }^{2}x\sin xdx$
$=\int (1-{\cos }^{2}x)\sin xdx$
Let $\cos x=t$, then $-\sin xdx=dt$.
$=\int (1-t^2)(-dt)$
$=\int (t^2-1)dt$
$=\frac{t^3}{3}-t+C$
$=\frac{{\cos }^{3}x}{3}-\cos x+C$