Question

The value of ${\int }_1^{37\pi }\frac{\pi \sin \left(\pi \log x\right)}{x}dx$

Answer 1

$={\int }_1^{37\pi }\frac{\pi \sin \left(\pi \log x\right)}{x}dx$

put $\pi logx=t$ and $\frac{\pi }{x}dx=dt$

$={\int }_1^{37\pi }\sin tdt$

now integrate

$={[-\cos t]}_1^{37\pi }$

put $\pi logx=t$

$={[-\cos (\pi \log x\left)\right]}_1^{37\pi }$

$=(-\cos (\pi \log \left(37\pi \right)\left)\right)-(-\cos \left(\pi \log 1\right))$

$=(-\cos (\pi \log \left(37\pi \right)\left)\right)-(-\cos \left(\pi \right(0\left)\right))$

$=(-\cos (\pi \log \left(37\pi \right)\left)\right)-(-\cos 0)$

$=(-\cos (\pi \log \left(37\pi \right)\left)\right)-(-1)$

$=(-\cos (\pi \log \left(37\pi \right)\left)\right)+1$

$=1-\cos (\pi \log \left(37\pi \right))$