If $\mathrm{I}=\int \frac{{\sec }^2\mathrm{x}-7}{{\sin }^7\mathrm{x}}\mathrm{dx},$ then I equals

$I=\int \frac{{\sec }^{2}x-7}{{\sin }^{7}x}dx$
$=\int \left({\sin }^{-7}x\right){\sec }^{2}xdx-\int 7{\mathrm{cosec}}^{7}xdx$
$$\begin{gathered} =\left({\sin }^{-7}x\right)\tan x-\int (-7){\sin }^{-8}x\cos x\cdot \tan xdx-7\int {\mathrm{cosec}}^{7}xdx \\ \left(by parts\right) \end{gathered}$$
$=\frac{\tan x}{{\sin }^{7}x}+7\int {\sin }^{-7}dx-7\int {\sin }^{-7}xdx$
$=\frac{\tan x}{{\sin }^{7}x}+c$