How do you integrate #int ln(sint)cost# by integration by parts method?

Question

Paisley Johnson · Accepted Answer

#intln(sint)costdt=sint(ln(sint)-1)+C#

First let #w=sint#, implying that #dw=costdt#. This will help simplify the problem before we attempt integration by parts. Through this substitution:

#I=intln(sint)costdt=intln(w)dw#

Now we should apply integration by parts. Let:

#{(u=ln(w)" "=>" "du=1/wdw),(dv=dw" "=>" "v=w):}#

Thus:

#I=wln(w)-intw1/wdw=wln(w)-intdw#

Integrating and factoring:

#I=wln(w)-w=w(ln(w)-1)#

From #w=sint#:

#I=sint(ln(sint)-1)+C#

Christopher Allers · Answer

undefined To integrate $ \int \ln(\sin(t)) \cos(t) \, dt $ using integration by parts, let $ u = \ln(\sin(t)) $ and $ dv = \cos(t) \, dt $. Then, $ du = \frac{\cos(t)}{\sin(t)} \, dt $ and $ v = \sin(t) $. Applying the integration by parts formula:

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

we have:

$$ \int \ln(\sin(t)) \cos(t) \, dt = \ln(\sin(t)) \sin(t) - \int \sin(t) \frac{\cos(t)}{\sin(t)} \, dt $$

$$ = \ln(\sin(t)) \sin(t) - \int \cot(t) \cos(t) \, dt $$

$$ = \ln(\sin(t)) \sin(t) - \int \frac{\cos^2(t)}{\sin(t)} \, dt $$

$$ = \ln(\sin(t)) \sin(t) + \int \frac{1 - \sin^2(t)}{\sin(t)} \, dt $$

$$ = \ln(\sin(t)) \sin(t) + \int \frac{1}{\sin(t)} \, dt - \int \sin(t) \, dt $$

$$ = \ln(\sin(t)) \sin(t) + \int \csc(t) \, dt - (-\cos(t)) + C $$

$$ = \ln(\sin(t)) \sin(t) - \ln|\csc(t) + \cot(t)| - \cos(t) + C $$

where $ C $ is the constant of integration.