Question

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

Answer 1

The answer is `I=(3/4)x[sin(ln(x))-cos(ln(x))]+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 `sinf(x)` or `cosf(x)`.

The theorem of integration by parts says:

`intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx`

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

For the first step it is necessary to find `g(x)=intg'(x)dx=int1dx=x`, and it is necessary to find `f'(x)=cos(ln(x))1/x`.

So:

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

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=3xsin(ln(x))-3xcos(ln(x))-3intsin(ln(x))dx=3xsin(ln(x))-3xcos(ln(x))-I`

because the last integral is identical to the first.

So:

`2I=3xsin(ln(x))-3xcos(ln(x))rArr`
`I=(3/2)x[sin(ln(x))-cos(ln(x))]+c`