Let g : 0 , ∞ → R be a differentiable function such that ∫ x cos x - sin x e x + 1 + g x e x + 1 - x e x e x + 1 2 d x = x g x e x + 1 + C , for all x > 0 , where C is an arbitrary constant. Then

Step-by-step solution

∫ x e x + 1 cos x - sin x d x + ∫ g x e x + 1 - x e x e x + 1 2 d x = x e x + 1 sin x + cos x - ∫ e x + 1 - x e x e x + 1 2 sin x + cos x d x + ∫ g x e x + 1 - x e x e x + 1 2 d x By comparison, we get, g x = sin x + cos x ⇒   g x = 2 sin x + π 4 Since x ∈ 0 , π 4 so, x + π 4 ∈ π 4 , π 2 So g x is increasing in 0 , π 4 g ' x = cos x - sin x i.e. g x - g ' x = 2 sin x is an increasing function in 0 , π 2 .