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Evaluate the following integrals:

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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