1. Integral sin(ln(x))dx

Find:

$\displaystyle \int \sin(\ln x) \,dx$

Substitution won't help, and integration by parts doesn't get seem to get you anywhere

Thanks.

2. Originally Posted by DivideBy0
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

3. 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:

$\displaystyle 2\int{sin(lnx)}dx=xsin(lnx)-xcos(lnx)$

$\displaystyle \boxed{\int{sin(lnx)}dx=\frac{xsin(lnx)}{2}-\frac{xcos(lnx)}{2}+C}$

4. Thanks, I thought parts wouldn't work because for some reason I was trying to integrate u instead of differentiate it. But i guess now it does work

5. Yes, sometimes it's easy to overlook what I call "cyclic IBP's". Those are the ones when you use parts you end up with the integral you started with. Then when you add it to both sides, you can divide by 2, 4, whatever and voila.