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Mathematics Question on Integrals of Some Particular Functions

If ∫f(x)log cos xdx=−log(log cos x)+C\int \frac{f\left(x\right)}{log\,cos\,x}dx=-log\left(log\,cos\,x\right)+C, then f(x)f\left(x\right) is equal to

A

tan xtan\,x

B

−sin x-sin\,x

C

−cos x-cos\,x

D

−tan x-tan\,x

Answer

tan xtan\,x

Explanation

Solution

∫f(x)log⁡ cos⁡ xdx=−log⁡(log⁡cos⁡ x)+C\int \frac{f(x)}{\log\, \cos\, x} dx=-\log (\log \cos\, x)+C
Differentiating on both sides w.r.t. xx,
\frac{d}{dx}\left\\{\int \frac{f(x)}{\log \,\cos\, x} d x\right\\}
=−ddxlog⁡(log⁡cos⁡x)+ddx(C)=-\frac{d}{d x}\\{\log (\log \cos x)\\}+\frac{d}{d x}(C)
⇒ f(x)log⁡cos⁡x=−1log⁡cos⁡x⋅1cos⁡x⋅(−sin⁡x)+0\Rightarrow \, \frac{f(x)}{\log \cos x}=\frac{-1}{\log \cos x} \cdot \frac{1}{\cos x} \cdot(-\sin x)+0
⇒ f(x)log⁡cos⁡x=tan⁡xlog⁡cos⁡x\Rightarrow\, \frac{f(x)}{\log \cos x}=\frac{\tan x}{\log \cos x}
⇒ f(x)=tan⁡ x\Rightarrow \, f (x) =\tan \,x