Question

$\int \left[\frac{1+\sin \left(\log x\right)}{1+\cos \left(\log x\right)}\right]dx=$

• A. $\frac{x}{1+\cos \left(\log x\right)}+c$
• B. $x\tan \frac{\log x}{2}+c$
• C. $-x\cot \frac{\log x}{2}+c$
• D. $\frac{x}{1+\sin \left(\log x\right)}+c$

Answer 1

The correct option is B $x\tan \frac{\log x}{2}+c$

$\int \left[\frac{1+\sin \left(\log x\right)}{1+\cos \left(\log x\right)}\right]dx$
$\frac{1}{2}\int \frac{1+\sin \left(\log x\right)}{{\cos }^2\left(\frac{\log x}{2}\right)}dx$
$=\int \left[\frac{1}{2}{\sin }^2\right(\frac{\log x}{2})+\tan (\frac{\log x}{2}\left)\right]dx$
It is in the form of
$\int \left(x{f}^1\right(x)+f(x\left)\right)dx=xf\left(x\right)+c$
$=x\tan \left(\frac{\log x}{2}\right)+c$
So, $\int \left[\frac{1+\sin \left(\log x\right)}{1+\cos \left(\log x\right)}\right]dx=x\tan \left(\frac{\log x}{2}\right)+c$