This function is surprisingly tricky, but integration by parts works. The trick is in finding the right parts! Recall that to integrate by parts we must divide the integrand into a product of 2 functions. But sin(ln(x)) doesn't seem to be factorable in this way.

But suppose we write du=dx, v=sin(ln(x)). It doesn't look too promising, but who knows? So our integrand, sin(ln(x))dx, is certainly v du. So let's get to work! Clearly we may take `u=x`, thus

(1) ∫ sin(ln(x))dx = ∫ v du = u*v - ∫ u dv =

= x sin(ln(x)) - ∫ x cos(ln(x)) 1/x dx = x sin(ln(x)) - ∫ cos(ln(x))dx,

and it looks like we've managed to reach the similar problem of integrating cos(ln(x)). But let's try the same trick again:

(2) ∫ cos(ln(x))dx =
= x cos(ln(x)) + ∫ x sin(ln(x)) 1/x dx =

(note how the

sign of the integral has flipped, since ∫ cos x dx = -sin(x)+C). Continuing,

= x cos(ln(x)) + ∫ sin(ln(x)) dx

Now just put (1) and (2) together:

>
∫ sin(ln(x)) dx = x sin(ln(x)) - x cos(ln(x)) - ∫ sin(ln(x)) dx,

so

>
2 ∫ sin(ln(x)) dx = x sin(ln(x)) - x cos(ln(x)) + C

and (dividing by 2) ∫ sin(ln(x)) dx = (x sin(ln(x)) - x cos(ln(x)))/2 + C.

It is also possible to perform the integration by using Euler's formulae for sin and cos in terms of complex exponentials; it is perhaps a somewhat harder method to justify on a term paper.