Evaluate the integral. (Use C for the constant of integration.) e4θ sin 5θ dθ?

NetherCraft 0

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

Also Check This  Which solution will decrease in volume?

Leave a Reply

Your email address will not be published.