Euler's log-trig integrals elegantly linked to a Berkeley Math Tournament integral

Exactly what the title says. You can solve these integrals using basic properties of the definite integral and high school integration techniques with some cool intuition.

[ThAlEdison]

Instead of arguing geometrically, I would've done another phase shift. For x on 0 to pi, if you replace it with pi/2-x, you get the range -pi/2 to pi/2. And the inside function becomes cos. Then you can use the property that even functions symmetric around the origin can be integrated by taking twice the value going from 0 to the upper bound.

[SabaSaa]

the integral of ln(1+sin(x)) over 0 to pi/2 also has a very nice result, but I cannot figure out how to do it.

[manstuckinabox]

We finally did it! we showed those IBP haters that they're opinions are very valid because I am one of them, but no seriously the berkley integral really yells by parts once you notice it's literally the integral of x*cot(x), now for the link I can see how there's a missing function for the berkley one to be the derivative function of something like ln(tanx)... say if it were for example sec^2/tan(x), but that's cos/sin*cos dx... which is 2 * cosx/sin(2x) which is so close to the derivative of ln(sin2x)... but I'm going to watch the actual vid and then see where's the link

I watched the video, and ooh mah lord that was awesome!

[rohanmadhuatMCK]

This question also came in my boards exam

[jieyuenlee]

Can be solve using contour intergration?

[sevdaniftaliyeva]

exactly for what do we consider the fact that the ln is a 1 to 1 function, is it to show that the ln(sinx) is also symmetric around pi/2?

[holyshit]

This is dummy variable because this is definite integral For the indefinite integral variable matters