Q.
Let g:(0,∞)→R be a differentiable function such that ∫(ex+1x(cosx−sinx)+(ex+1)2g(x)(ex+1−xex))dx=ex+1xg(x)+c
for all x>0, where c is an arbitrary constant. Then.
∫(ex+1x(cosx−sinx)+(ex+1)2g(x)(ex+1−xex))dx=ex+1xg(x)+c
On differentiating both sides w.r.t. x, we get (ex+1x(cosx−sinx)+(ex+1)2g(x)(ex+1−xex) =(ex+1)2(ex+1)(g(x)+xg′(x))−ex⋅x⋅g(x) (ex+1)x(cosx−sinx)+g(x)(ex+1−xex) =(ex+1)(g(x)+xg′(x))−ex⋅x⋅g(x) ⇒g′(x)=cosx−sinx ⇒g(x)=sinx+cosx+C g(x) is increasing in (0,π/4) g′′(x)=−sinx−cosx<0 ⇒g′(x) is decreasing function
let h(x)=g(x)+g′(x)=2cosx+C ⇒h′(x)=g′(x)+g′′(x)=−2sinx<0 ⇒h is decreasing
let ϕ(x)=g(x)−g′(x)=2sinx+C ⇒ϕ′(x)=g′(x)−g′′(x)=2cosx>0 ⇒ϕ is increasing