Can anyone help me find the Integral. ∫ (dt/t*√(t^2-64))?

Question

i just want to know what i should let t equal

Answer 1

The antiderivative of 1/(t Sqrt[t^2 - 64]) is equal to

ArcSec[t/8]/8 + C,

where C is an arbitrary constant. The method of integration is by trigonometric substitution. Let t = 8 Sec[u], so that dt = 8 Sec[u]Tan[u] du. Then Sqrt[t^2 - 64] = 8 Sqrt[Sec[u]^2 - 1] = 8 Tan[u], and it follows that the transformed integrand is

8 Sec[u] Tan[u]/(8 Sec[u] 8 Tan[u]) du = 1/8 du,

which integrates to u/8 + C, and after substituting back the value of t, we obtain the desired result.

Answer 2

In general, if the integrand is a function of x and sqrt(x^2 - a^2), use the substitution

u = inv sec (x/a)
x = a sec u
sqrt(x^2 - a^2) = a tan u
dx = sec u tan u du

In your case, x = t, a = 8, and you get

INT 1/[t sqrt(t^2-64)] dt = INT 1/[64 sec u tan u] sec u tan u du

= INT du/8 = u/8 + C

with back substitution

= inv sec (t/8) / 8 + C = inv cos(8/t) / 8 + C

Answer 3

∫ (dt/t*√(t^2-64)) = 0.5 ∫ (dt^2/t^2√(t^2-64)) =
set u = t^2 plug and play
0.5 ∫ (du/u√(u-64)) = split fractions eyc,

Answer 4

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