Help me solve this intergral

6S4 1/(((t^2) - 9) ^ (1/2)) dt

integrate 1 over the square root of t squared - 9 ....from 4 to 6

im using u sub and i get 36(sqrt3) - 8(sqrt7) , its wrong and the real answer has a natural log, not sure what to do

Hello,

6
∫ [1 /√(t² - 9)] dt =
4

rewrite the integral as:

∫ [1 /√(t² - 3²)] dt =

let:

t = 3secθ ↔ secθ = t/3

dt = 3tanθ secθ dθ

then, substituting:

∫ [1 /√(t² - 3²)] dt = ∫ {1 /√[(3secθ)² - 3²]} 3tanθ secθ dθ =

∫ [1 /√(3²sec²θ - 3²)] 3tanθ secθ dθ =

∫ {1 /√[3²(sec²θ - 1)]} 3tanθ secθ dθ =

replace sec²θ - 1 with tan²θ:

∫ [1 /√(3²tan²θ)] 3tanθ secθ dθ =

∫ [1 /(3tanθ)] 3tanθ secθ dθ =

simplifying into:

∫ secθ dθ =

multiply the integrand by (tanθ + secθ) /(secθ + tanθ) (= 1):

∫ (tanθ + secθ) secθ dθ /(secθ + tanθ) =

(expanding)

∫ (tanθ secθ + sec²θ) dθ /(secθ + tanθ) =

note that we have in the numerator the derivative of the denominator:

∫ d(secθ + tanθ) /(secθ + tanθ) =

(being this of the form ∫ d[f(x)] /f(x) = ln |f(x)| + C)

ln |secθ + tanθ| + C

recall that secθ = t/3

hence:

tanθ = √(sec²θ - 1) = √[(t/3)² - 1] = √[(t²/9) - 1] = √[(t² - 9)/9] = [√(t² - 9)] /3

then, substituting back:

ln |secθ + tanθ| + C = ln |(t /3) + {[√(t² - 9)] /3}| + C =

ln |[t + √(t² - 9)] /3| + C =

(applying logarithm properties)

ln |t + √(t² - 9)| - ln (3) + C

being - ln (3) a mere constant, it can be included in C, so the antiderivative is:

ln |t + √(t² - 9)| + C

having the antiderivative, let's plug in the bounds:

ln |6 + √(6² - 9)| - ln |4 + √(4² - 9)| =

ln |6 + √(36 - 9)| - ln |4 + √(16 - 9)| =

ln (6 + √27) - ln (4 + √7) =

ln (6 + 3√3) - ln (4 + √7) =

(applying logarithm properties)

ln [(6 + 3√3) /(4 + √7)]


the answer is: ln [(6 + 3√3) /(4 + √7)] (≈ 0.52159)


I hope it helps