I'm confused on how to solve or set up a substitute for the problem.
Evaluate the Integral from 1/4 to 1.
1/ sqrt(16t^2 -1) dt
To evaluate the given integral: ∫(1/√(16t^2 - 1)) dt, you can utilize the substitution method. Follow these steps:
Step 1: Let u = 16t^2 - 1.
By making this substitution, you can transform the integral into a simpler form.
Step 2: Find the derivative of u with respect to t, du/dt.
du/dt = d/dt (16t^2 - 1)
= 32t
Step 3: Rearrange the differential equation in terms of dt.
Divide both sides of the equation by 32t:
dt = (1/32t) du
Step 4: Substitute the new variables and limits into the integral.
When t = 1/4, u = 16(1/4)^2 - 1 = 4 - 1 = 3.
When t = 1, u = 16(1)^2 - 1 = 16 - 1 = 15.
Now, rewrite the integral using the new variables:
∫(1/√(16t^2 - 1)) dt = ∫(1/√u) (1/32t) du
Step 5: Simplify the integral.
Since dt = (1/32t) du, substitute dt with the given expression:
∫(1/√u) (1/32t) du = (1/32) ∫(1/√u) du
Step 6: Evaluate the integral with respect to u.
Integrating (1/√u)du yields 2√u + C, where C is the constant of integration.
Step 7: Substitute back the original variable.
Substituting back u = 16t^2 - 1, we have:
2√(16t^2 - 1) + C.
Step 8: Evaluate the integral with the given limits.
Now, evaluate the integral with the limits t = 1/4 and t = 1:
[2√(16(1)^2 - 1) + C] - [2√(16(1/4)^2 - 1) + C]
Simplifying further, we get:
2√15 + C - 2√3 + C
Step 9: Combine like terms and present the final answer.
The final answer is 2√15 - 2√3 + C, where C represents the constant of integration.