Prove that :
$$\int^{\pi/2}_0 \log|\sin x| \,dx = \int^{\pi/2}_0 \log|\cos x| \,dx $$
I tried to cut the integral into a sum of parts and changing variable but it didn't work out right, i dont know how to solve this kind of problems in any other way, any hint will be much appreciated!
Hint: Use the fact $$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$ You can show this by letting $u = a-x$.
Set $t = \pi/2-x$ and make use of the fact that $\displaystyle \int_a^b f(y) dy = - \int_b^a f(y) dy$ to conclude what you want.
$$I=\int_0^{\dfrac{\pi}{2}}\log\sin x\,dx\cdots \tag 1 $$ use this property:$$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$ $$I=\int_0^{\dfrac{\pi}{2}}\log\cos x\,dx\cdots \tag 2$$
Add both eqn: $$2I=\int_0^{\dfrac{\pi}{2}}\log\sin x+\log\cos x\,dx$$ $$2I=\int_0^{\dfrac{\pi}{2}}\log({\sin x\cdot\cos x})\,dx$$ $$2I=\int_0^{\dfrac{\pi}{2}}\log(\dfrac{2\sin x\cdot\cos x}{2})\,dx$$ $$2I=\int_0^{\dfrac{\pi}{2}}(\log\sin 2x-\log2)\,dx$$ $$2I=\int_0^{\dfrac{\pi}{2}}\log\sin 2x\,dx-\int_0^{\dfrac{\pi}{2}}\log2\,dx$$ In first integration put $2x=t$ $$2I=\dfrac12\int_0^\pi{\log\sin t}\,dt-\dfrac {\pi}{2}\log 2$$ use this property: If $f(2a-x)=f(x)$ then $$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$$ $$2I=\dfrac22\int_0^\dfrac{\pi}{2}{\log\sin t}\,dt-\dfrac {\pi}{2}\log 2$$ $$2I=\int_0^\dfrac{\pi}{2}{\log\sin x}\,dx-\dfrac {\pi}{2}\log 2$$ $$2I=I-\dfrac {\pi}{2}\log 2$$ $$I=-\dfrac {\pi}{2}\log 2$$ $$I=\int_0^{\dfrac{\pi}{2}}\log\sin x\,dx=\int_0^{\dfrac{\pi}{2}}\log\cos x\,dx=-\dfrac {\pi}{2}\log 2$$
