int_0^pilog(sin^2 x) dx=0?
1 Answer
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
Make a substitution
I=ln(4)pi+4int_0^(pi/2) ln(sin(u))du+4int_0^(pi/2) ln(cos(u))du
Make a substitution
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
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