Question

What is `int sin(lnx) dx`?

Answer 1

Use integration by parts twice and solve for the integral to find

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

Explanation:

We will proceed through integration by parts.

Let `I = intsin(ln(x))dx`

Integration by Parts (1)

Let `u = sin(ln(x))` and `dv = dx`

Then `du = cos(ln(x))/xdx` and `v = x`

Applying the integration by parts formula `intudv = uv - intvdu`

`I = intsin(ln(x))dx`

`= xsin(ln(x)) - intcos(ln(x))dx`

Integration by Parts (2)

Let `u = cos(ln(x))` and `dv = dx`

Then `du = -sin(ln(x))/x` and `v = x`

Applying the formula, we have

`intcos(ln(x))dx = xcos(ln(x)) - int-sin(ln(x))dx`

`= xcos(ln(x)) + intsin(ln(x))dx`

`= xcos(ln(x)) + I`

Substituting this into the result from the first integration by parts, we obtain

`I = xsin(ln(x)) - xcos(ln(x)) - I`

`=> 2I = x(sin(ln(x)) - cos(ln(x)))`

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

But in the process of adding `I` to both sides, we lost the constant, and so for our final answer, we add it back in to get

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