How do you integrate #int sin(lnx)# using integration by parts?

Question

Christopher Allers · Accepted Answer

#1/2([xsin(ln(x))]-[xcos(ln(x))])#

#intsin(ln(x))dx#

Let's #u = ln(x)#

#du = dx/x#

#dx=xdu#

#x=e^u#

then

#inte^usin(u)du#

By part : #dv = e^u ; v = e^u ; w = sin(u) ; dw = cos(u) #

#[v*w]-intdw*v#

#inte^usin(u)du=[e^usin(u)]-inte^ucos(u)du#

Once more, by part

#dv=e^u ; v = e^u ; w = cos(u) ; dw = -sin(u)#

#inte^usin(u)du=[e^usin(u)]-([e^ucos(u)]+inte^usin(u)du)#

#inte^usin(u)du=[e^usin(u)]-[e^ucos(u)]-inte^usin(u)du#

#2inte^usin(u)du=[e^usin(u)]-[e^ucos(u)]#

#inte^usin(u)du=1/2([e^usin(u)]-[e^ucos(u)])#

Substitute back for #u = ln(x)#

#intsin(ln(x))dx=1/2([xsin(ln(x))]-[xcos(ln(x))])#

Aurora Alvarado · Answer

undefined To integrate $ \int \sin(\ln(x)) \, dx $ using integration by parts, let $ u = \ln(x) $ and $ dv = \sin(\ln(x)) \, dx $. Then, differentiate $ u $ to find $ du $, and integrate $ dv $ to find $ v $. Next, apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute $ u $, $ dv $, $ du $, and $ v $ into the formula and evaluate the integral.

$$ u = \ln(x) $$
$$ dv = \sin(\ln(x)) \, dx $$

Differentiating $ u $:

$$ du = \frac{1}{x} \, dx $$

Integrating $ dv $:

$$ v = -\cos(\ln(x)) $$

Now, apply the integration by parts formula:

$$ \int \sin(\ln(x)) \, dx = -\ln(x)\cos(\ln(x)) - \int (-\cos(\ln(x))) \frac{1}{x} \, dx $$

Simplify:

$$ \int \sin(\ln(x)) \, dx = -\ln(x)\cos(\ln(x)) + \int \frac{\cos(\ln(x))}{x} \, dx $$

Now, integrate $ \int \frac{\cos(\ln(x))}{x} \, dx $ using substitution or integration by parts again, depending on your preference and convenience.