Question:
Evaluate the following integrals:
$\int \frac{\sin (2+3 \log x)}{x} d x$
Solution:
Assume $2+3 \log x=t$
$\mathrm{d}(2+3 \log \mathrm{x})=\mathrm{dt}$
$\Rightarrow \frac{3}{x} d x=d t$
$\Rightarrow \frac{1}{x} d x=\frac{d t}{3}$
Substituting $t$ and $d t$
$\Rightarrow \frac{1}{3} \int \sin t d t$
$=-\cos t+c$
But $t=2+3 \log x$
$\Rightarrow \frac{-1}{3} \cos (2+3 \log x)+c$
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