Question:
Evaluate the following integrals:
$\int \cos (\log x) d x$
Solution:
Let $\mathrm{I}=\int \cos (\log \mathrm{x}) \mathrm{dx}$
Let $\log x=t$
$\frac{1}{x} d x=d t$
$d x=x d t$
$=\int e^{t} \cos t d t$
We know that, $\int \cos (\log x) d x=\frac{e^{\operatorname{ax}}}{a^{2}+b^{2}}\{a \sin (b x+c)-b \cos (b x+c)\}$
Hence, $a=1, b=1$
So, $\mathrm{I}=\frac{\mathrm{e}^{\mathrm{t}}}{2}[\cos \mathrm{t}+\sin \mathrm{t}]+\mathrm{c}$
Hence,
$\int \cos (\log x) d x=\frac{e^{\log x}}{2}\{\cos (\log x)+\sin (\log x)\}+c$
$I=\frac{x}{2}\{\cos (\log x)+\sin (\log x)\}+c$
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