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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

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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 $f(x) = - \cot(x)$.

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