Trigonometric/Logarithmic Integration Can you please help me find this integral?
$$\int \sin(\ln(x)) dx$$
Give me a clue or show step by step solutions please.
Thank you very much.
 A: Putting $\ln x=y, x=e^y, dx=e^y dy$
So, $\int \sin(\ln x)dx=\int \sin y\cdot e^y dy$
Use Integration by parts, with $e^y$ as the first term

Alternatively,  using Euler's formula, $e^{iy}=\cos y+i\sin y$ 
$\int \sin y\cdot e^y dy$ is the imaginary part of $\int e^y\cdot e^{iy}dy$
$$\int e^y\cdot e^{iy}dy=\int e^{y(1+i)}dy=\frac{e^y(e^{iy})}{(1+i)}=\frac{(1-i)e^y(\cos y+i\sin y)}2=\frac{e^y\{(\cos y+\sin y)+i(\sin y-\cos y)\}}2$$
$$\implies \int \sin y\cdot e^y dy=\frac{e^y(\sin y-\cos y)}2$$
$$\implies \int \sin(\ln x)dx=\frac{x(\sin(\ln x)-\cos(\ln x))}2$$
A: Make a substitution:  $u = \ln x$.  Then $du = \frac{1}{x} dx$, so $dx = x du$.  Then you can use the fact that $e^u = x$.  
Hope this helps!
A: Let our integral be $I$. Use integration by parts. Let $u=\sin(\ln x)$ and $dv=dx$. Then $u=\frac{1}{x}\cos(\ln x)$, and we can take $v=x$. Thus 
$$I=x\cos(\ln x)-\int \cos(\ln x)\,dx.$$
Let $J=\int \cos(\ln x)\,dx$. The same sort of calculation as the one above yields
$$J=-x\cos(\ln x)+\int \sin(\ln x)\,dx.$$
Thus 
$$I=x\cos(\ln x)-J\qquad\text{and}\qquad J=-x\sin(\ln x)+I.$$
Solve for $I$, and don't forget the $+C$. We get
$$I=\frac{x\cos(\ln x)+x\ln(\sin x)}{2} +C.$$
