$\int \sin ^5 x \mathrm{~d} x=$

To evaluate the integral $\int \sin^5 x \, dx$, the substitution method is employed by expressing $\sin^5 x$ as $\sin^4 x \cdot \sin x$ and using the identity $\sin^2 x = 1 - \cos^2 x$.

$\int (1 - \cos^2 x)^2 \cdot \sin x \, dx$

Let $u = \cos x$, so $du = -\sin x \, dx$ and $\sin x \, dx = -du$.

Substituting gives $-\int (1 - u^2)^2 \, du$.

Expanding the integrand: $-\int (1 - 2u^2 + u^4) \, du$.

Integrating term by term: $-\left(u - \frac{2u^3}{3} + \frac{u^5}{5}\right) + c$.

Replacing $u = \cos x$ yields $-\left(\cos x - \frac{2\cos^3 x}{3} + \frac{\cos^5 x}{5}\right) + c$.

Distributing the negative sign gives the final answer:

$-\cos x + \frac{2\cos^3 x}{3} - \frac{\cos^5 x}{5} + c$

Assuming option C contains a minor typographical error ($\cos^2 x$ should be $\cos^3 x$), it corresponds to the derived result.