I = ∫ π 4 π 3 8 sin x - sin 2 x x d x . Then

Step-by-step solution

Let, f x = 8 sin x - sin 2 x ⇒ f ' x = 8 cos x - 2 cos 2 x ⇒ f ' ' x = - 8 sin x + 4 sin 2 x ⇒ f ' ' x = - 8 sin x 1 - cos x So, f ' x < 0   when  x ∈ π 4 , π 3 Hence   f ' x is decreasing function in given domain, Now, f ' π 3 < f ' x < f ' π 4 ⇒ 5 < f ' x < 8 2 ⇒ 5 < f ' x < 4 2 Now integrating all with respect to d x ∫ 5 d x < ∫ f ' x d x < ∫