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$\int \sin x . \log (\sec x + \tan x) d x = f (x) + x + c$ then f(x)=

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$- \cos x . \log (\sec x + \tan x)$

$\sin x . \log (\sec x + \tan x)$

$- \cos x . logsec x$

$\cos x . \log (\sec x + \tan x)$

answer is C.

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Detailed Solution

$\int \log (secx + t a n x) . \sin x d x$

$= \log (\sec x + \tan x) (- \cos x) - \int \frac{1 (\sec x \tan x + \sec^{2} x)}{\sec x + \tan x} (- \cos x) d x$

$= (- \cos x) \log (\sec x + \tan x) + \int \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} (+ \cos x) d x$

$= (- \cos x) \log (\sec x + \tan x) + x + c$

$f (x) = - \cos x . l o g (\sec x + \tan x)$

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