Evaluate the following integrals:

Assume $1+(\log x)^{2}=t$

$d\left(1+(\log x)^{2}\right)=d t$

$\Rightarrow \frac{2 \log x}{x} d x=d t$

$\Rightarrow \frac{\log x}{x} d x=\frac{d t}{2}$

$\Rightarrow \int \sin t \frac{d t}{2}$

$\Rightarrow \frac{1}{2} \int \sin t d t$

$\Rightarrow \frac{-1}{2} \cos t+c$

But $t=1+(\log x)^{2}$

$\Rightarrow \frac{-1}{2} \cos \left(1+\log x^{2}\right)+c$