I need help solving Integral from 0 to (a/sqrt(b)) of (a^2-bx^2)^(3/2) I got (a^4/8sqrt(b)) - 3a^4pi/4sqrt(b)
Assistance needed.
SDSU = ??
∫(a^2 - bx^2) dx
As a test, I let a=2 and b=3 and ran it through
the Wolfram integrator.
Looks at the mess I got with nice numbers....
I have idea.
Do you have a table of integrals?
See if you can find your pattern.
http://integrals.wolfram.com/index.jsp?expr=%284-3x%5E2%29%5E%283%2F2%29&random=false
To solve the integral ∫[0, a/√b] (a^2 - bx^2)^(3/2), we can use a combination of trigonometric and hyperbolic substitutions. Let's go step by step:
Step 1: Make a substitution
Let's perform the substitution x = (a/√b)sinh(t). By doing this, we can convert the integral into a form that can be solved easily.
dx = (a/√b)cosh(t)dt
x = (a/√b)sinh(t) could be rewritten as sinh(t) = (√b/a)x
Step 2: Substitute the integral limits
When x = 0, t = 0. And when x = (a/√b), we need to determine the value of t.
Using the substitution in the equation:
sinh(t) = (√b/a) * (a/√b) = 1
This implies t = sinh^(-1)(1)
Step 3: Convert the expression inside the integral
Now we need to express (a^2 - bx^2)^(3/2) in terms of t.
(a^2 - bx^2)^(3/2) = [(a^2 - b(a^2/bsinh^2(t)))]^(3/2)
= [(a^2 - a^2cosh^2(t))]^(3/2)
= (a^2(1 - cosh^2(t)))^(3/2)
= a^3(sinh^2(t))^(3/2)
= a^3sinh^3(t)
Step 4: Express the integral with the substitution
Now that we have converted the integral, we can rewrite it using the substitution.
∫[0, a/√b] (a^2 - bx^2)^(3/2) dx
= ∫[0, sinh^(-1)(1)] a^3sinh^3(t) (a/√b)cosh(t) dt
= (a^4/√b) ∫[0, sinh^(-1)(1)] sinh^3(t)cosh(t) dt
Step 5: Simplify the integral
To simplify the integration, we can use the identity cosh^2(t) - sinh^2(t) = 1.
(a^4/√b) ∫[0, sinh^(-1)(1)] sinh^3(t)cosh(t) dt
= (a^4/√b) ∫[0, sinh^(-1)(1)] [sinh^2(t) - sinh^4(t)]cosh(t) dt
= (a^4/√b) ∫[0, sinh^(-1)(1)] [1 - cosh^2(t)][sinh^2(t)cosh(t)] dt
Step 6: Substitution
Let's substitute u = sinh(t), then du = cosh(t) dt.
(a^4/√b) ∫[0, sinh^(-1)(1)] [1 - u^2]u du
Step 7: Evaluate the integral
Integrate the expression using the power rule for integration.
(a^4/√b) ∫[0, sinh^(-1)(1)] (u - u^3) du
= (a^4/√b) [(u^2/2) - (u^4/4)] [0, sinh^(-1)(1)]
= (a^4/√b) [((sinh^(-1)(1))^2/2) - ((sinh^(-1)(1))^4/4)]
Step 8: Final Simplification
Finally, simplify the expression and substitute back the original variables.
(a^4/8√b) - (3a^4π/32√b)