Given expression is
xx∫0(1−t)sin(f(t))dt=2x∫0tsin(f(t))dt
Differentiating w.r.t. x using Leibnitz theorem, we get
x∫0(1−t)sin[f(t)]dt+x(1−x)sin[f(x)]=2xsin[f(x)]
⇒x∫0(1−t)sin[f(t)]dt=xsin[f(x)](1+x)
Again differentiating w.r.t. x, we get
(1−x)sin[f(x)]=(x+x2)cos[f(x)]f′(x)+(2x+1)sin[f(x)]
⇒−3xsin[f(x)]=(x+x2)cos[f(x)]f′(x)
⇒−3xx(1+x)=cot[f(x)]f′(x)
⇒f′(x)cotf(x)+31+x=0