If ∫1−5sin2⁡xcos5⁡xsin2⁡xdx=f(x)cos5⁡x+C, then f(x):

∫1−5sin2⁡xcos5⁡xsin2⁡xdx=∫cosec2⁡xcos5⁡xdx−5∫1cos5⁡xdx=−cot⁡xcos5⁡x+∫cot⁡x−5cos−6⁡x(−sin⁡x)dx−5∫1cos5⁡xdx=−cot⁡xcos5⁡x+5∫dxcos5⁡x−5∫dxcos5⁡x+C=−cot⁡xcos5⁡x+CSo f(x)=-cotx

If ∫1−5sin2⁡xcos5⁡xsin2⁡xdx=f(x)cos5⁡x+C, then f(x):

∫1−5sin2⁡xcos5⁡xsin2⁡xdx=∫cosec2⁡xcos5⁡xdx−5∫1cos5⁡xdx=−cot⁡xcos5⁡x+∫cot⁡x−5cos−6⁡x(−sin⁡x)dx−5∫1cos5⁡xdx=−cot⁡xcos5⁡x+5∫dxcos5⁡x−5∫dxcos5⁡x+C=−cot⁡xcos5⁡x+CSo f(x)=-cotx

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If $\int \frac{1 - 5 \sin^{2} x}{\cos^{5} x \sin^{2} x} d x = \frac{f (x)}{\cos^{5} x} + C$ , then $f (x) :$

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$- \cos e c x$

$c o t x$

$\cos e c x$

$- c o t x$

answer is D.

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Detailed Solution

$\begin{array}{l} \int \frac{1 - 5 \sin^{2} x}{\cos^{5} x \sin^{2} x} d x = \int \frac{{cosec}^{2} x}{\cos^{5} x} d x - 5 \int \frac{1}{\cos^{5} x} d x \\ = - \frac{\cot x}{\cos^{5} x} + \int \cot x (- 5 \cos^{- 6} x) (- \sin x) d x - 5 \int \frac{1}{\cos^{5} x} d x \\ = - \frac{\cot x}{\cos^{5} x} + 5 \int \frac{d x}{\cos^{5} x} - 5 \int \frac{d x}{\cos^{5} x} + C \\ = - \frac{\cot x}{\cos^{5} x} + C \end{array}$
So $f (x) = - c o t x$