We can calculate it with substitutions Substitution u = sin^2(x) will give us 1/4\int_{0}^{1}\frac{ln{u}}{\sqrt{1-u^2}}du then substitution u = sin(t) gives us integral 1/4\int_{0}^{\frac{\pi}{2}}\ln{sin{u}}du
👍 I = - (1/2) ln (cosec x)^2/√{1+(cosec x)^2 } let (cosec x)^2 = cosec t d x = cot t dt/2√ {(cosec t - 1) } I = - (1/4) integral of ln cosec t in [0,π/2] I = - (π/8) ln 2
We can calculate it with substitutions
Substitution u = sin^2(x) will give us
1/4\int_{0}^{1}\frac{ln{u}}{\sqrt{1-u^2}}du
then substitution u = sin(t)
gives us integral 1/4\int_{0}^{\frac{\pi}{2}}\ln{sin{u}}du
That is beautiful and much simpler.
👍
I = - (1/2) ln (cosec x)^2/√{1+(cosec x)^2 }
let (cosec x)^2 = cosec t
d x = cot t dt/2√ {(cosec t - 1) }
I = - (1/4) integral of ln cosec t in [0,π/2]
I = - (π/8) ln 2
This approach is slightly different then mine but essentially the same
done