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