Question

Solve $\int \sec \left(\log x\right)[1+\tan (\log x\left)\right]dx$

• A. $x\sec \left(\log x\right)+c$
• B. $\frac{x}{2}\sec \left(\log x\right)+c$
• C. $-x\sec \left(\log x\right)+c$
• D. $\frac{-x}{2}\sec \left(\log x\right)+c$

Answer 1

The correct option is A $x\sec \left(\log x\right)+c$

$\int sec\left(logx\right)[1+tan(logx\left)\right]dx$

$=\int \left[sec\right(logx)+sec(logx\left)tan\right(logx\left)\right]dx$

$=\int sec\left(logx\right)dx+\int sec\left(logx\right)tan\left(logx\right)dx$

We know, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

Now by integrating the functions by parts, for the first part, taking $sec\left(logx\right)$ as $u$ and $x$ as $v$, we get

$=xsec\left(logx\right)-\int \frac{sec\left(logx\right)tan\left(logx\right)}{x}\cdot xdx+\int sec\left(logx\right)tan\left(logx\right)dx$

$=xsec\left(logx\right)+c$

Hence, option 'A' is correct.