In this section, we will find the integration of log(sin x) from 0 to pi/2. The following property will be used to calculate the integral.
$\int_0^af(x)dx=\int_0^af(a-x)dx$ …(i)
Integration of log(sin x) from 0 to pi/2
Question: Show that
$\int_0^{\frac{\pi}{2}} \log \sin x dx = – \frac{\pi}{2} \log 2$
Solution:
Let $I=\int_0^{\frac{\pi}{2}} \log \sin x dx$ …(ii)
Using the above property (i), we have
$I=\int_0^{\frac{\pi}{2}}\log \sin xdx$ $=\int_0^{\frac{\pi}{2}}\log \sin(\frac{\pi}{2}-x)dx$
By the fact $\sin(\frac{\pi}{2}-x)=\cos x$ we get that
$I=\int_0^{\frac{\pi}{2}}\log \cos x dx$ ...(iii)
Adding (ii) and (iii), we obtain that
$2I=\int_0^{\frac{\pi}{2}}(\log \sin x+\log \cos x)dx$
$\Rightarrow 2I=\int_0^{\frac{\pi}{2}}\log(\sin x \cos x)dx$ as log(a) + log(b)=log(ab)
$\Rightarrow 2I=\int_0^{\frac{\pi}{2}}\log(\frac{\sin 2x}{2})dx$ as sin(2x)=2sin x cos x
$\Rightarrow 2I=\int_0^{\frac{\pi}{2}}(\log \sin 2x-\log 2)dx$ as log(a) – log(b)=log(a/b)
$\Rightarrow 2I=\int_0^{\frac{\pi}{2}}\log \sin 2x dx$ $- \int_0^{\frac{\pi}{2}} \log 2dx$
$\Rightarrow 2I = \int_0^{\frac{\pi}{2}}\log \sin 2x dx-\frac{\pi}{2}\log 2$ …(iv)
Let $I_1=\int_0^{\frac{\pi}{2}}\log \sin 2xdx$
Put $t=2x$
Therefore, dt=2 dx. Thus dx=dt/2.
Note that $t=\pi$ when $x=\frac{\pi}{2}$ and $t=0$ when $x=0$
So $I_1=\frac{1}{2}\int_0^\pi\log(\sin t) dt$ …(v)
Property: $\int_0^{2a}f(x)dx=2\int_0^a f(a-x)dx$ when $f(2a-x)=f(x)$
Observe that we have $\log(\sin t)=\log \sin(pi-t)$
Thus, by the above property
$I_1=\frac{1}{2} \cdot 2 \int_0^{\frac{\pi}{2}} \log(\sin t) dt$
$=\int_0^{\frac{\pi}{2}} \log(\sin t) dt=I$
As $I_1=I$ we deduce from (iv) and (v) that
$2I =I-(\frac{\pi}{2})\log 2$
$\Rightarrow I=-\frac{\pi}{2} \log 2$
So we finally get that
$\int_0^{\frac{\pi}{2}} \log \sin x=-\frac{\pi}{2}\log 2$