Question $14$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{3}xdx=$

$\frac{co{s}^{3}x}{3}-x+C$

$\frac{si{n}^{4}x}{4}+C$

$-\frac{co{s}^{3}x}{3}+cosx+C$

$\frac{co{s}^{3}x}{3}-cosx+C$