If f(x)=cos⁡x and g(x)=sin⁡x then ∫log⁡f(x)(f(x))2dx is

I=∫log⁡cos⁡xcos2⁡xdx=∫sec2⁡xlog⁡cos⁡xdx=(tan⁡x)log⁡cos⁡x−∫tan⁡x−sin⁡xcos⁡xdx=(tan⁡x)log⁡cos⁡x+∫sec2−1dx=(tan⁡x)[log⁡cos⁡x+1]−x+C=g(x)f(x)[log⁡f(x)+1]−x+C

If f(x)=cos⁡x and g(x)=sin⁡x then ∫log⁡f(x)(f(x))2dx is

I=∫log⁡cos⁡xcos2⁡xdx=∫sec2⁡xlog⁡cos⁡xdx=(tan⁡x)log⁡cos⁡x−∫tan⁡x−sin⁡xcos⁡xdx=(tan⁡x)log⁡cos⁡x+∫sec2−1dx=(tan⁡x)[log⁡cos⁡x+1]−x+C=g(x)f(x)[log⁡f(x)+1]−x+C

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If $f (x) = \cos x$ and $g (x) = \sin x$ then $\int \frac{\log f (x)}{(f (x))^{2}} d x$ is

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$\frac{g (x)}{f (x)} (\log f (x) + 1) + \frac{x^{2}}{2} + C$

$f (x) (\log f (x) + 1) + x + C$

$\frac{f (x)}{g (x)} (\log g (x) + 1) + x + C$

$\frac{g (x)}{f (x)} [\log f (x) + 1] - x + C$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int \frac{\log \cos x}{\cos^{2} x} d x = \int \sec^{2} x \log \cos x d x \\ = (\tan x) \log \cos x - \int \tan x (\frac{- \sin x}{\cos x}) d x \\ = (\tan x) \log \cos x + \int (\sec^{2} - 1) d x \\ = (\tan x) [\log \cos x + 1] - x + C \end{array}$