Question

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

• A. $-\sin x$
• B. $\cos x$
• C. $\tan x$
• D. $\cot x$
  

Answer 1

The correct option is A $-\sin x$

$\int cosxlog\left(cosx\right)dx$

$-log\left(cosx\right)\int cosxdx$

$-\int \left(\frac{1}{cosx}\right(-sinx)\int cosxdx)dx$

$=\left(sinx\right)log\left(cosx\right)+\int \frac{si{n}^{2}x}{cosx}dx$

$=sinxlog\left(cosx\right)+\int secxdx-\int cosxdx+c$

$sinxlog\left(cosx\right)+log|secx+tanx|-sinx+c$

so, $f\left(x\right)=-sinx$