Question

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

Answer 1

`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