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
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
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