Question

How do you evaluate the integral `int tanthetaln(sintheta)`?

Answer 1

`-1/2[ln(sin(x))ln(sin(x)+1)+Li_2(-sin(x))-Li_2(1-sin(x))]`

Explanation:

We will begin by using the `tan(phi)=sin(phi)/cos(phi)` identity to rewrite the integral:
`int\ tan(x)ln(sin(x))\ dx=int\ sin(x)/cos(x)ln(sin(x))\ dx`

Next we will multiply on the top and bottom by `cos(x)` so we can use the Pythagorean identity:
`int\ cos(x)((sin(x)ln(sin(x)))/cos^2(x))\ dx`

`=int\ cos(x)((sin(x)ln(sin(x)))/(1-sin^2(x)))\ dx`

Now we can let `u=sin(x)` and divide by `(du)/dx=cos(x)`
`=int\ cancel(cos(x)/cos(x))((sin(x)ln(sin(x)))/(1-sin^2(x)))\ du`

`=int\ (u\ln(u))/(1-u^2)\ du`

Now we need to do partial fractions. I will bring out a `-1` out of the bottom to make factoring easier:
`=int\ (u\ln(u))/-(-1+u^2)\ du=-int\ (u\ln(u))/(u^2-1)`

`=-int\ (u\ln(u))/((u+1)(u-1))\ du`

Now we do partial fractions:
`(u\ln(u))/((u+1)(u-1))=A/(u+1)+B/(u-1)`

After multiplying by the left hand side denominator, we're left with:
`u\ln(u)=A(u-1)+B(u+1)`

If we expand, setup an equation system and solve, we get:
`A=B=1/2ln(u)`

Now our integral has become:
`-int\ (u\ln(u))/(u^2-1)\ du=-1/2(int\ ln(u)/(u+1)\ du+int\ ln(u)/(u-1)\ du)`

I will call the left one Integral 1 and the right one Integral 2.

Integral 1
Here we have to do integration by parts with `f=ln(u)` and `g'=1/(u+1)`

We know `f'=1/u` and `g=ln|u+1|`, so we get:
`int\ ln(u)/(u+1)\ du=ln(u)ln|u+1|-int\ ln(u+1)/u\ du`

I will call this rightmost integral Integral 3

Integral 3
This integral has no elementary solution, but we see that it is relatively close to the form for the dilogarithm, which looks like this:
`Li_2(t)=int-ln(1-t)/t\ dt`

If we introduce a substitution, `z=-u` and `(dz)/(du)=-1`, we can get the integral to this form:
`-int\ ln(1-z)/-z\ dz=-int-ln(1-z)/z\ dz=-Li_2(z)`

If we resubstitute, we get that Integral 3 is equal to:
`-Li_2(-u)`

Completing Integral 1
We have evaluated Integral 3, so if we plug in, we get:
`int\ ln(u)/(u+1)\ du=ln(u)ln|u+1|+Li_2(-u)`

Integral 2
This integral can also be reduced into the dilogarithm. This time we will introduce a substitution with `z=u-1`:
`int\ ln(u)/(u-1)\ du=int\ ln(z+1)/z\ dz`

This is the same as Integral 3, so we get:
`int\ ln(z+1)/z\ dz=-Li_2(-z)=-Li_2(1-u)`

Completing the original integral
Now that we have evaluated Integral 1 and Integral 2, we can combine them to get our final answer:
`-int\ (u\ln(u))/(u^2-1)\ du=-1/2(ln(u)ln|u+1|+Li_2(-u)-Li_2(1-u))`

If we resubstitute and see that `sin(x)+1` is never negative, so we can remove the absolute value, we have:
`-1/2[ln(sin(x))ln(sin(x)+1)+Li_2(-sin(x))-Li_2(1-sin(x))]`