TITLE: integral of $\sin(\ln(x))\,dx$
QUESTION [5 upvotes]: I tried to calculate this integral:  $$\int \sin(\ln(x))\,dx$$ but it seems my result is wrong. My calculation is:
 $$\int  \sin(\ln(x)) \,dx = \left| \begin{array}{c} u=\ln x \\  du=(1/x)dx\\  
dx=xdu \end{array}  \right|$$
with integration by parts:
 $$-\cos(u)x-\int -\cos(u) dx$$
 $$-\cos(u)x+\sin(u)$$
 $$\sin(ln|x|)-\cos(\ln|x|)x +C$$ 
The answer in the website is:
$$0.5[\sin(\ln(x))x-\cos(\ln(x))x]$$
What did I do wrong?

REPLY [2 votes]: \begin{align}
\int dx \sin\ln x &= \int du e^u \sin u\\
&= \int du \dfrac{e^{(1+i)u} - e^{(1-i) u}}{2i}\\
&= \dfrac{e^u}{2i} \left(\dfrac{e^{(1+i)u}}{1+i} - \dfrac{e^{(1-i)u}}{1-i}\right)\\
&= \dfrac{e^u}{2i} \left(\dfrac{(1-i)e^{iu} - (1+i) e^{-i u}}{2} \right)\\
&= \dfrac{e^u}{2i} \left(\dfrac{e^{iu} - e^{-iu}}{2} -i \dfrac{e^{iu} + e^{-iu}}{2} \right)\\
&=\dfrac{e^u}{2} \left(\sin u - \cos u \right)\\
&= \dfrac{x}{2}\left(\sin\ln x - \cos\ln x\right)
\end{align}