Question

$\int\left(sec\,x\right)log \left(sec\,x-tan\,x\right)dx=$

• A. \frac{1}{2}\left\\{log\left(sec\,x-tan\,x\right)\right\\}^{2}+C
• B. -\frac{1}{2}\left\\{log\left(sec\,x-tan\,x\right)\right\\}^{2}+C
• C. -\frac{3}{2}\left\\{log\left(sec\,x-tan\,x\right)\right\\}^{2}+C
• D. -\frac{1}{2}\left\\{log\left(sec\,x-tan\,x\right)\right\\}^{-2}+C

Answer 1

Answer: (B)

-\frac{1}{2}\left\\{log\left(sec\,x-tan\,x\right)\right\\}^{2}+C

Answer 2

Let $I = \int sec\,x \cdot log(sec\,x - tan\, x)dx$
Put $log(sec\,x - tan\,x) = t$
$\Rightarrow \frac{1}{sec\,x - tan\,x}\times (sec\,x \,tan\,x - sec^2\,x)dx = dt$
$\Rightarrow - sec\,x\,dx = dt$
$\therefore I = -\int t \,dt = \frac{-t^2}{2} + C$
$= \frac{-1}{2}[log(sec\,x - tan\,x)]^2 + C$