Evaluate the integral. ∫ (√(x^2 - 16))/x^3 dx?
I have tried this problem a couple times and I still haven't gotten it right. I am not sure what I am doing wrong. Can you write out the steps and the answer. This is just a practice problem for homework. Thank you!
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Let x = 4secϴ
4/x = cosϴ → ϴ= arccos(4/x)
dx = 4secϴtanϴ dϴ
x² = 16sec²ϴ
and
x³ = 64sec³ϴ
Substituting all the x's and dx with ϴ:
∫ (√(16sec²ϴ - 16) / 64sec³ϴ · 4secϴtanϴ dϴ
= ∫ (√(16(sec²ϴ - 1)) / 16sec²ϴ · tanϴ dϴ
= ∫ (√(16tan²ϴ) / 16sec²ϴ · tanϴ dϴ
= ∫4tanϴ / 16sec²ϴ · tanϴ dϴ
= ¼ ∫tan²ϴ / sec²ϴ dϴ
= ¼ ∫sin²ϴ dϴ
= ¼ ∫½ - ½cos(2ϴ) dϴ
= ⅛ϴ - (1/16)sin(2ϴ) + C
= ⅛ϴ - ⅛sinϴcosϴ + C
= ⅛·arccos(4/x) - ⅛·√(x²-16)/x ·(4/x)+ C
= ⅛·arccos(4/x) - ½√(x²-16)/x² + C
â« (â(x^2 - 16))/x^3 dx
Let x = 4 sec t
dx = 4 sec t tan t dt
x^2-16 = 16 sec^2 t - 16 = 16(sec^2 t-1) = 16 tan^2 t
sqrt(x^2-16) = 4 tan t
â« (â(x^2 - 16))/x^3 dx = â« 4 tan t [4 sec t tan t] dt / 64 sec^3 t
= (1/4) â« tan^2 t / sec^2 t dt
= (1/4) â« sin^2 t dt
= (1/4) â« (1- cos 2t) /2 dt
= (1/8) â« 1 dt - (1/8) â« cos 2t dt
= (1/8) t - (1/8) sin 2t (1/2)
= (1/8) t - (1/8) (2 sin t cos t)(1/2)
= (1/8) t - (1/8) sin t cos t --------- (1)
Our substitution was x = 4 sec t
x = 4 / cos t
cos t = 4/x
sin t = sqrt(1-cos^2 t) = sqrt(1- 16/x^2) = sqrt(x^2-16) / x
(1) becomes:
(1/8) sec^-1(x/4) - (1/8) sqrt(1-16/x^2) (4/x)
= (1/8) sec^-1(x/4) - sqrt(1-16/x^2) / 2x
= (1/8) sec^-1(x/4) - sqrt(x^2-16) / 2x^2 + C
â«â(x^2 - 16)/x^3 dx
Let x = 4sec(t) => dx = 4sec(t)tan(t) dt
â«16tan(t) * 4sec(t)tan(t) dt/64sec^3(t) dt
â«tan^2(t)/sec^2(t) dt
= 1/2 * â«(1 - cos(2t)) dt
= t/2 - 1/2*sin(t)cos(t) + C
= 1/2*arcsec(x/4) - 2â(x^2 - 16)/x^2 + C