Home > Calculus > Integration by Parts

How do you evaluate the integral #int sinthetaln(costheta)#?

Answer 1

Isaiah Allen

#int sin theta ln (cos theta) d theta = -costheta[ln(cos theta) -1] +C#

If we substitute:

#t=cos theta# #dt = -sin theta d theta#

we have that:

#int sin theta ln (cos theta) d theta = -int lnt dt#

We can calculate this integral by parts:

#int lnt dt = tlnt -int t d(lnt) =tlnt - int dt = tlnt-t +C#

So, substituting back #theta# we have:

#int sin theta ln (cos theta) d theta = -costheta(ln(cos theta) -1) +C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Isaiah Allen

To evaluate the integral ∫ sin(θ) ln(cos(θ)) dθ, we can use integration by parts.

Let u = ln(cos(θ)) and dv = sin(θ) dθ. Then, du = -tan(θ) dθ and v = -cos(θ).

Applying the integration by parts formula: ∫ u dv = uv - ∫ v du,

we get: ∫ sin(θ) ln(cos(θ)) dθ = -ln(cos(θ)) * cos(θ) - ∫ -cos(θ) * (-tan(θ)) dθ.

Simplify and integrate the remaining integral: = -cos(θ) ln(cos(θ)) + ∫ cos(θ) tan(θ) dθ.

Now, we have another integral to evaluate, which can be done by using substitution method or recognizing it as a known integral.

After evaluating the integral, you will obtain the final result.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.