# int_0^pilog(sin^2 x) dx=0?

Apr 12, 2018

$I = - \ln \left(4\right) \pi$

#### Explanation:

We want to solve

$I = {\int}_{0}^{\pi} \ln \left({\sin}^{2} \left(x\right)\right) \mathrm{dx}$

By the properties of logarithms

$I = 2 {\int}_{0}^{\pi} \ln \left(\sin \left(x\right)\right) \mathrm{dx}$

Use the trig identity

color(blue)(sin(x)=2sin(x/2)cos(x/2)

Thus

$I = 2 {\int}_{0}^{\pi} \ln \left(2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)\right) \mathrm{dx}$

$= 2 {\int}_{0}^{\pi} \ln \left(2\right) \mathrm{dx} + 2 {\int}_{0}^{\pi} \ln \left(\sin \left(\frac{x}{2}\right)\right) \mathrm{dx} + 2 {\int}_{0}^{\pi} \ln \left(\cos \left(\frac{x}{2}\right)\right) \mathrm{dx}$

$= \ln \left(4\right) \pi + 2 {\int}_{0}^{\pi} \ln \left(\sin \left(\frac{x}{2}\right)\right) \mathrm{dx} + 2 {\int}_{0}^{\pi} \ln \left(\cos \left(\frac{x}{2}\right)\right) \mathrm{dx}$

Make a substitution color(red)(u=x/2=>du=1/2dx
color(red)(x=0=>u=0 and color(red)(x=pi=>u=pi/2

$I = \ln \left(4\right) \pi + 4 {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(u\right)\right) \mathrm{du} + 4 {\int}_{0}^{\frac{\pi}{2}} \ln \left(\cos \left(u\right)\right) \mathrm{du}$

Make a substitution color(red)(u=-pi/2-s=>du=-ds
color(red)(u=0=>s=-pi/2 and color(red)(u=pi/2=>s=-pi

$I = \ln \left(4\right) \pi + 4 {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(u\right)\right) \mathrm{du} - 4 {\int}_{- \frac{\pi}{2}}^{- \pi} \ln \left(- \sin \left(s\right)\right) \mathrm{ds}$

$\textcolor{w h i t e}{I} = \ln \left(4\right) \pi + 4 {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(u\right)\right) \mathrm{du} - 4 {\int}_{- \frac{\pi}{2}}^{- \pi} \ln \left(\sin \left(- s\right)\right) \mathrm{ds}$

Make a substitution color(red)(w=-s=>dw=-ds
color(red)(s=-pi/2=>w=pi/2 and color(red)(s=-pi=>w=pi

$I = \ln \left(4\right) \pi + 4 {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(u\right)\right) \mathrm{du} + 4 {\int}_{\frac{\pi}{2}}^{\pi} \ln \left(\sin \left(w\right)\right) \mathrm{dw}$

$I = \ln \left(4\right) \pi + 2 I$

$I = - \ln \left(4\right) \pi$

A very neat result