how come if you change from pi to pi/2, the integral is doubled? I get that if the integrand was just sinx but isn't lnsinx something entirely different?This is basically a trick. There is no elementary primitive. Here's a clue. Change the range of integration to 0 to pi/2 and call the integral I. Then you want -2*I. Now observe the integral of log(cos(x)) is also I. That's the clue. Add integral log(sin(x)) and log(cos(x)) and use a rule of logarithms and a trig identity and a u-substitution. Now you got an equation with a bunch of I's in it. Can you solve for I?
I only solved it based on your hints but I didn't know it was symmetric at pi/2. But how you knew to change the limits of integration beats meThey are very different. But they are still symmetric around x=pi/2. Did you solve the problem? It's really not that hard if you put your mind to it and know the secret hint. I only knew it because I've seen this problem before.
Surely, you didn't get to where you are being expect to solve problems like this without learning that the derivative of cos x is -sin x??i can do it but i need the primitive of sin x and thats my problem, the rest of the problem i can solve , i just only cant manage to discover de primitive of sin x
We already answered this below!Surely, you didn't get to where you are being expect to solve problems like this without learning that the derivative of cos x is -sin x??