Answer
$$\frac{1}{3}\cos^3x-3\cos x-3\sec x+\frac{1}{3}\sec^3x +c$$
Work Step by Step
\begin{align*}
\int \frac{\sin ^{7} x} {\cos ^{4} x}d x&= \int \sin^{6}x\cos^{-4}x\sin xdx\\
&= \int \cos^{-4}x(1-\cos^2x)^3\sin xdx\\
&=\int \cos^{-4}x(-\cos^6x+3\cos^4x-3\cos^2x+1 )\sin xdx\\
&= \int \left( -\cos^2x+3-3\cos^{-2}x+\cos^{-4}x \right)\sin xdx\\
&= \frac{1}{3}\cos^3x-3\cos x-3\sec x+\frac{1}{3}\sec^3x +c
\end{align*}
