If $\int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin^2 x}dx = \frac{f(x)}{\cos^5 x}+\mathcal {C}$. Then value of $f(x)$. If  $\displaystyle \int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin^2 x}dx = \frac{f(x)}{\cos^5 x}+\mathcal {C}$. Then value of $f(x)$.
$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin ^2 x}dx = \underbrace{\int \frac{\csc^2 x}{\cos^5 x}dx}_{=I}-5\int\frac{1}{\cos^5 x}dx$$
So $$\displaystyle I = \int\frac{1}{\cos ^5 x}\cdot \csc^2 xdx = \frac{1}{\cos^5 x}\cdot -\cot x+5\int\frac{-\sin x}{\cos^6 x}\cdot -\cot xdx$$
$$\displaystyle  = -\frac{\cot x}{\cos^5 x}+5\int\frac{1}{\cos^5 x}dx+\mathcal{C}$$
So $$\displaystyle \int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin^2 x}dx = -\frac{\cot x}{\cos^5 x}+\mathcal{C}=\frac{f(x)}{\cos^5 x}+\mathcal{C}$$
So $f(x) = -\cot x$.
Can we solve it any other Method, If Yes then plz explain here
Thanks 

Consider the differentiation of both sides of the integral to find $f(x)$. In this view differentiating 
\begin{align}
\int \frac{1-5 \sin^{2}(x)}{\cos^{5}(x) \, \sin^{2}(x)} \, dx = \frac{f(x)}{\cos^{5}(x)} + c
\end{align}
leads to
\begin{align}
f'(x) + 5 \tan(x) \, f(x) = \csc^{2}(x) -5.
\end{align}
Now, the right-hand side of this equation can be written in such a way that
\begin{align}
f'(x) + 5 \tan(x) \, f(x) &= \frac{d}{dx}( - \cot(x)) + 5 \tan(x) \, (-\cot(x)) \\
\left[ \frac{d}{dx} + 5 \tan(x) \right] f(x) &= \left[ \frac{d}{dx} + 5 \tan(x) \right](-\cot(x))
\end{align}
which yields that $f(x) = - \cot(x)$. 
