Question

How do I solve `int_0^1sin(x)/(sin(x)+sin(1-x))dx`?

`int_0^1sin(x)/(sin(x)+sin(1-x))dx`
I tried to solve this by making a u-substitution, where `u=1-x`.
`u=1-x`
`x=1-u`
`int_0^1sin(1-u)/(sin(1-u)+sin(u))dx`
But now, I'm stuck again. I'm not sure what to do from here.

Answer 1

`I=1/2`
NOTE: `I=int_0^1sin(1-u)/(sin(1-u)+sin(u))du=int_0^1sin(1-x)/(sin(1-x)+sin(x))dx`
Then adding this with given `equ^n`,we get,`I+I=2I` and so on...

Explanation:

We note that.`color(red)((1)I=int_a^bf(x)dx=int_a^bf(t)dt=>)`,
The value of `I` is not dependent on variable `tox,t,u...`etc.
`color(red)((2)I=int_0^af(x)dx=int_0^af(a-x)dx,)` If f is continuous on`[0,a].`
`I=int_0^1(sinx)/(sinx+sin(1-x))dx.........to (A)`
Applying above theorem (2) we get,`I=int_0^1(sin(1-x))/(sin(1-x)+sin(x))dx.......to (B)`
Adding `(A) and (B)` ,i.e .`(A)+(B)=>I+I=2I`
`2I=int_0^1(sinx)/(sinx+sin(1-x))dx+int_0^1(sin(1-x))/(sin(1-x)+sin(x))dx`Denominator of both integrals are same.
`2I=int_0^1(sinx+sin(1-x))/(sinx+sin(1-x))dx=int_0^1(1)dx=[x]_0^1=1-0`
`2I=1=>I=1/2`