Answer
$$-\left(\cos x-\frac{1}{3}\cos^3x+\frac{1}{5}\cos ^5x\right) +C$$
Work Step by Step
$$\int \sin^5x dx$$
Use $$\sin^5x= \sin^4x \sin x= (1-\cos^2 x)^2\sin x$$
\begin{align*}
\int \sin^5x dx&=\int (1-\cos^2 x)^2\sin xdx,\ \ \\
\text{Let }\ u=\cos x \, \ du=-\sin xdx\\
&= -\int (1-2u^2+u^4)du\\
&=-\left(u-\frac{1}{3}u^3+\frac{1}{5}u^5\right) +C\\
&= -\left(\cos x-\frac{1}{3}\cos^3x+\frac{1}{5}\cos ^5x\right) +C
\end{align*}
