# How do I evaluate int3 sin(ln(x))dx?

Jan 27, 2015

The answer is $I = \left(\frac{3}{4}\right) x \left[\sin \left(\ln \left(x\right)\right) - \cos \left(\ln \left(x\right)\right)\right] + c$

This integral has to be done two times by parts because it is a cyclic integral. A cyclic integral is an integral of a cyclic function, like $\sin f \left(x\right)$ or $\cos f \left(x\right)$.

The theorem of integration by parts says:

$\int f \left(x\right) g ' \left(x\right) \mathrm{dx} = f \left(x\right) g \left(x\right) - \int g \left(x\right) f ' \left(x\right) \mathrm{dx}$

In this case, it seems that there is only one function, but it is not true because we can assume that $f \left(x\right) = \sin \left(\ln \left(x\right)\right)$ and $g ' \left(x\right) \mathrm{dx} = 1 \mathrm{dx}$.

For the first step it is necessary to find $g \left(x\right) = \int g ' \left(x\right) \mathrm{dx} = \int 1 \mathrm{dx} = x$, and it is necessary to find $f ' \left(x\right) = \cos \left(\ln \left(x\right)\right) \frac{1}{x}$.

So:

$I = \int 3 \sin \left(\ln \left(x\right)\right) \mathrm{dx} = 3 \int \sin \left(\ln \left(x\right)\right) \mathrm{dx} = 3 \left[x \sin \left(\ln \left(x\right)\right) - \int \cos \left(\ln \left(x\right)\right) \left(\frac{1}{x}\right) x \mathrm{dx}\right] = 3 \left[x \sin \left(\ln \left(x\right)\right) - \int \cos \left(\ln \left(x\right)\right) \mathrm{dx}\right] =$

Now it is necessary to make another time the integration by parts:

3[xsin(ln(x))-(xcos(ln(x))-int-sin(ln(x))(1/x)xdx]=3[xsin(ln(x))-xcos(ln(x))-intsin(ln(x))dx]

So:

$I = 3 x \sin \left(\ln \left(x\right)\right) - 3 x \cos \left(\ln \left(x\right)\right) - 3 \int \sin \left(\ln \left(x\right)\right) \mathrm{dx} = 3 x \sin \left(\ln \left(x\right)\right) - 3 x \cos \left(\ln \left(x\right)\right) - I$

because the last integral is identical to the first.

So:

$2 I = 3 x \sin \left(\ln \left(x\right)\right) - 3 x \cos \left(\ln \left(x\right)\right) \Rightarrow$
$I = \left(\frac{3}{2}\right) x \left[\sin \left(\ln \left(x\right)\right) - \cos \left(\ln \left(x\right)\right)\right] + c$