Intergrate sin(ln(x)) dx?

Question

Should I do thiis by parts? or how do I go about this

Answer 1

u = sin(ln(x))
du = cos(ln(x)) * 1/x dx
dv = dx
v = x

∫ u dv = uv - ∫v du
sin(ln(x))*x - ∫ x * [cos(ln x)]/x dx
x*sin(ln(x)) - ∫ cos(ln x) dx

Now repeat for ∫ cos(ln x) dx
u = cos(ln(x))
du = - sin(ln(x))/x
dv = dx
v = x

x*cos(ln(x)) - ∫x*(-sin(ln(x))/x) dx
x*cos(ln(x)) + ∫ sin(ln(x)) dx


Putting everything back we have:
∫ sin(ln(x)) dx = x*sin(ln(x)) - [x*cos(ln(x)) + ∫ sin(ln(x)) dx]
Notice how the original integral shows up in the answer.

Integral = x*sin(ln(x)) - x*cos(ln(x)) - Integral
2 Integral = x*sin(ln(x)) - x*cos(ln(x))

∫ sin(ln(x)) dx = [x*sin(ln(x)) - x*cos(ln(x))] / 2 + C

Answer 2

Let's try by parts. Put u = sin(ln(x)) and dv = dx.
Then, Int sin(ln(x)) dx = x sin(ln(x) - Int x cos(ln(x) *1/x dx =>
Int sin(ln(x)) dx = x sin(ln(x) - Int cos(ln(x) dx

Now, again by parts,

Int sin(ln(x)) dx = x sin(ln(x) - [x cos(ln(x)) - Int x (-sin(ln(x)) * 1/x dx]. So,

Int sin(ln(x)) dx = x sin(ln(x) - x cos(ln(x)) - Int sin(ln(x)) dx

So, we get the same integral on the Right hand side. Therefore, 2 Int sin(ln(x)) dx = x[sin(ln(x)) cos(lnx))] + 2C, C the integration constant. And finally,

Int sin(ln(x)) dx = x[sin(ln(x)) cos(lnx))]/2 + C

Answer 3

It's tricky, but parts works.

Let du=dx and v=sin(ln(x))

Then u=x and dv=cos(ln(x))/x dx

So via the formula (uv-int(udv)) we get

xsin(ln(x))-int (cos(ln(x)))dx

Now if we use parts on int (cos(ln(x))), we get

xcos(ln(x))+int(sin(ln(x)))dx

We put everything together and we get

int(sin(ln(x))dx)
=xsin(ln(x))-xcos(ln(x))
-int(sin(ln(x))dx)

So we bring the integral from the right side to the left side and get

2int(sin(ln(x))dx)
=xsin(ln(x))-xcos(ln(x))

So int(sin(ln(x))dx)

=[xsin(ln(x))-xcos(ln(x))]/2 +C