If `xint_0^xsin(f(t))dt=(x+2)int_0^x tsin(f(t))dt ,w h e r ex >0,` then show that `f^(prime)(x)cotf(x)+3/(1+x)=0.`

Explanation

Given equation is
`x int_(0)^(x)(1-t)sin(t(t))dt=2int_(0)^(x)tsin(f(t))dt`
Differentiating w.r.t `x` we get
`int_(0)^(x)(1-t)sin[f(t)]dt+x(1-x)sin[f(x)]=2x sin[f(x)]`
or `int_(0)^(x)(1-t)sin[f(t)]dt=x^(2)sin[(f(x))]+xsin[f(x)]`
Again, differentiating w.r.t `x`, we get
`(1-x)sin[(f(x))=2xsin[f(x)]+x^(2)cos[f(x)=f[(x)`
`+sin[f(x)]+x cos [f(x)]f[(x)`
or `-3xsin[f(x)]=(x+x^(2))cos[f(x)]f'(x)`
or `(-3x)/(x(1+x))=cot[f(x)]f'(x)`
or `f'(x)cotf(x)+3/(1+x)=0`