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.

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