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

1 Answer
Apr 12, 2018

I=-ln(4)pi

Explanation:

We want to solve

I=int_0^piln(sin^2(x))dx

By the properties of logarithms

I=2int_0^piln(sin(x))dx

Use the trig identity

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

Thus

I=2int_0^piln(2sin(x/2)cos(x/2))dx

=2int_0^piln(2)dx+2int_0^piln(sin(x/2))dx+2int_0^piln(cos(x/2))dx

=ln(4)pi+2int_0^piln(sin(x/2))dx+2int_0^piln(cos(x/2))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(4)pi+4int_0^(pi/2) ln(sin(u))du+4int_0^(pi/2) ln(cos(u))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(4)pi+4int_0^(pi/2) ln(sin(u))du-4int_(-pi/2)^(-pi) ln(-sin(s))ds

color(white)(I)=ln(4)pi+4int_0^(pi/2) ln(sin(u))du-4int_(-pi/2)^(-pi) ln(sin(-s))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(4)pi+4int_0^(pi/2) ln(sin(u))du+4int_(pi/2)^(pi) ln(sin(w))dw

I=ln(4)pi+2I

I=-ln(4)pi

A very neat result