Solve the integral

$\int_1^e \frac \partial {\partial x}\sin (\ln |x|) \ dx = \sin (\ln|x|) \ \bigg|_0^e = \sin (\ln e) - \sin (\ln 1) = \sin 1 - \sin 0 = \sin 1 \approx \boxed{0.841}$

First of all, you can cancel this derivative with the integration, and just apply the limits into the derivativent. $\sin(\ln e)-\sin(\ln 1) = \sin 1 - \sin 0$ , then you calculate the results $\sin 1 \approx 0.841471$ .

If $\displaystyle f=f(x)$ , then $\displaystyle df=\frac{\partial f}{\partial x} dx$ .

Hence, $\displaystyle \int_1^{e} \frac{\partial }{\partial x}\sin{(\ln{|x|})} dx =\int_1^{e} d \left(\sin{(\ln{|x|})}\right)$

According to FTOC II, we finally obtain

$\displaystyle \int_1^{e} \frac{\partial }{\partial x}\sin{(\ln{|x|})} dx =\int_1^{e} d \left(\sin{(\ln{|x|})}\right) = \sin{(\ln{|e|})} - \sin{(\ln{|1|})} = \sin{1}\approx 0.841$