Question

How do you integrate `sin( ln x )`?

Answer 1

`I = x/2 ( sin(ln x) - cos(ln x) )+C`

Explanation:

`I = int \ sin (ln x) \ dx`

this is in the IBP section meaning you don't really have much choice how to take this, so ...

`I =int \ d/dx(x) * sin (ln x) \ dx`

which by IBP

`= x sin(ln x) - int \ x *d/dx( sin (ln x)) \ dx`

`= x sin(ln x) - int \ x cos (ln x)* 1/x \ dx`

`= x sin(ln x) - int \ cos (ln x) \ dx`

and now another round of IBP

`= x sin(ln x) - int \ d/dx(x) * cos (ln x) \ dx`

`= x sin(ln x) - x cos(ln x) + int \ x d/dx( cos (ln x)) \ dx`

`= x sin(ln x) - x cos(ln x) + int \ x (-sin (ln x) 1/x) \ dx`

`= x sin(ln x) - x cos(ln x) - int \ sin (ln x) \ dx`

`= x sin(ln x) - x cos(ln x) - I`

`2I = x sin(ln x) - x cos(ln x) +C` making sure to add in the integration constant

`implies I = x/2 ( sin(ln x) - cos(ln x) )+C`

Answer 2

`int sin(log_e x) = x/2(sin(log_ex)-cos(log_ex))`

Explanation:

`e^{i phi} = cos phi+ i sin phi` (Moivre's identity)

so

`e^{i log_e x} = cos(log_e x)+i sin(log_e x)`

but

`e^{i log_e x} = (e^{log_e x})^i = x^i`

and

`int x^i dx = 1/2(1-i)x^{1+i} = x/2(1-i)x^i`

Substituting for `x^i` and taking the imaginary component we get

`int sin(log_e x) = x/2(sin(log_ex)-cos(log_ex))`