Question

$\int \sin x\log (\sec x+\tan x)dx=f\left(x\right)+x+c$ then $f\left(x\right)=$

• A. $\cos x\log (\sec x+\tan x)+c$
• B. $\sin x\log (\sec x+\tan x)+c$
• C. $-\cos x\log \sec x+\tan x)+c$
• D. $-\cos x\log \sec x+c$

Answer 1

The correct option is C $-\cos x\log \sec x+\tan x)+c$

$\int sinxlog(secx+\tan x)dx$
$=log(secx+\tan x)\int sinxdx$
$-\int \left(\frac{d}{dx}\right(log(secx+\tan x))\int sinxdx)dx$
$=-log(secx+\tan x)cosx+\int secxcosxdx$
$=-cosxlog(secx+\tan x)+x+c$
So, $f\left(x\right)=-cosxlog(secx+\tan x)$