Find:
$\displaystyle \int \sin(\ln x) \,dx$
Substitution won't help, and integration by parts doesn't get seem to get you anywhere
Thanks.
ah, but a combination works!
Note that, $\displaystyle \int \sin ( \ln x )~dx = \int \frac xx \sin ( \ln x )~dx$
Let $\displaystyle u = \ln x \implies x = e^u$
$\displaystyle \Rightarrow du = \frac 1x~dx$
So our integral becomes:
$\displaystyle \int e^u \sin u~du$
and integration by parts works on that guy
Also: i haven't tried this, but maybe using series can work. do you know the power series for sin(x)? perhaps you could do term by term integration
Parts will work. You just have to apply it twice.
Let $\displaystyle u=sin(lnx), \;\ dv=dx, \;\ du=\frac{cos(lnx)}{x}, \;\ v=x$
Then you get:
$\displaystyle xsin(lnx)-\int{cos(lnx)}dx$
Now, for the integral on the right:
Let $\displaystyle u=cos(lnx), \;\ dv=dx, \;\ du=\frac{-sin(lnx)}{x}dx, \;\ v=x$
Then we get:
$\displaystyle \int{sin(lnx)}dx=xsin(lnx)-xcos(lnx)-\int{sin(lnx)}dx$
Add $\displaystyle \int{sin(lnx)}dx$ to both sides:
$\displaystyle 2\int{sin(lnx)}dx=xsin(lnx)-xcos(lnx)$
$\displaystyle \boxed{\int{sin(lnx)}dx=\frac{xsin(lnx)}{2}-\frac{xcos(lnx)}{2}+C}$