Question:
Evaluate the following integrals:
$\int \log x \frac{\sin \left\{1+(\log x)^{2}\right\}}{x} d x$
Solution:
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 $t=1+(\log x)^{2}$
$\Rightarrow \frac{-1}{2} \cos \left(1+\log x^{2}\right)+c$
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