$\int^{\pi/2}_{0}\log|\sin x| \,dx = \int^{\pi/2}_{0}\log|\cos x| \,dx $ 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!
 A: Hint:  Use the fact $$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$  You can show this by letting $u = a-x$.
A: 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.
A: $$I=\int_0^{\dfrac{\pi}{2}}\log\sin x\,dx\cdots \tag 1 $$
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
$$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$$
$$2I=\dfrac22\int_0^\dfrac{\pi}{2}{\log\sin t}\,dt-\dfrac {\pi}{2}\log 2$$
$$2I=\int_0^\dfrac{\pi}{2}{\log\sin x}\,dx-\dfrac {\pi}{2}\log 2$$
$$2I=I-\dfrac {\pi}{2}\log 2$$
$$I=-\dfrac {\pi}{2}\log 2$$
$$I=\int_0^{\dfrac{\pi}{2}}\log\sin x\,dx=\int_0^{\dfrac{\pi}{2}}\log\cos x\,dx=-\dfrac {\pi}{2}\log 2$$
