$\int[\sin |\log x|+\cos |\log x|] d x=$

Question

Aicharya · Accepted Answer

Let $\mathrm{I}=\int[\sin |\log \mathrm{x}|+\cos |\log \mathrm{x}|] \mathrm{dx}$ Put $\log \mathrm{x}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}$ $\begin{aligned} & 	herefore \mathrm{I}=\int(\sin +\cos t) \mathrm{e}^{\mathrm{t}} d t \ & =\int \mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}+\cos t) d t=\mathrm{e}^{\mathrm{t}} \sin \mathrm{t}+\mathrm{c}=\mathrm{x} \sin |\log \mathrm{x}|+\mathrm{c} \end{aligned}$

$\int[\sin |\log x|+\cos |\log x|] d x=$

Solution