1 Answer

Let I = ∫ e^4θ sin 5θ dt
Integrate by parts
dv= e^4θ dθ ; v = (1/4) e^4θ
u = sin 5θ ; du = 5 cos 5θ
∫ u dv = u v – ∫ v du
I = (1/4) sin 5θ e^4θ – (5/4) ∫ e^4θ cos 5θ dθ ———— (1)
From (1), integrate ∫ e^4θ cos 5θ dθ by parts
dv = e^4θ dθ ; v = (1/4) e^4θ
u = cos 5θ ; du = 5 sin5θ dθ
∫ u dv = u v – ∫ v du
∫ e^4θ cos 5θ dθ = (1/4) cos 5θ e^4θ – ∫ (1/4) e^4θ ( 5 sin5θ dθ )
∫ e^4θ cos 5θ dθ = (1/4) cos 5θ e^4θ + (5/4) ∫ e^4θ sin5θ dθ
∫ e^4θ cos 5θ dθ = (1/4) cos 5θ e^4θ + (5/4) I
substitute this into (1)
I = (1/4) sin 5θ e^4θ – (5/4) ( (1/4) cos 5θ e^4θ + (5/4) I )
I = (1/4) sin 5θ e^4θ – (5/16) e^4θ cos 5θ – (25/16) I
I ( 1+ 25/16) = (1/4) sin 5θ e^4θ – (5/16) e^4θ cos 5θ
(41/16) I = (1/4) sin 5θ e^4θ – (5/16) e^4θ cos 5θ
I = (16/41)(1/4) sin 5θ e^4θ – (16/41) (5/16) e^4θ cos 5θ
I = (4/41) sin 5θ e^4θ – (5/41) cos 5θ e^4θ
∫ e^4θ sin 5θ dt = (4/41) sin 5θ e^4θ – (5/41) cos 5θ e^4θ + C