Math Problem Statement

Evaluate the integral \(\int_0^2 \sqrt{x(4-x)} \, dx\) using the substitution \(x = 4\sin^2\theta\), giving the answer in exact form.

Solution

Let's solve the integral 02x(4x)dx\int_0^2 \sqrt{x(4-x)} \, dx using the substitution x=4sin2θx = 4\sin^2\theta.

Step 1: Substitution

Given x=4sin2θx = 4\sin^2\theta, we need to calculate the corresponding expressions for dxdx, the limits of integration, and simplify the integrand:

  1. Substitute x=4sin2θx = 4\sin^2\theta:

    • Then, dx=8sinθcosθdθ=4sin(2θ)dθdx = 8\sin\theta\cos\theta \, d\theta = 4\sin(2\theta) \, d\theta (using the double-angle identity sin(2θ)=2sinθcosθ\sin(2\theta) = 2\sin\theta\cos\theta).
  2. Change the limits of integration:

    • When x=0x = 0, 4sin2θ=0    θ=04\sin^2\theta = 0 \implies \theta = 0.
    • When x=2x = 2, 4sin2θ=2    sin2θ=12    θ=π44\sin^2\theta = 2 \implies \sin^2\theta = \frac{1}{2} \implies \theta = \frac{\pi}{4}.

Step 2: Simplify the integrand

Now, we substitute into the integrand: x(4x)=4sin2θ(44sin2θ)=16sin2θcos2θ=4sinθcosθ=2sin(2θ)\sqrt{x(4-x)} = \sqrt{4\sin^2\theta(4 - 4\sin^2\theta)} = \sqrt{16\sin^2\theta\cos^2\theta} = 4\sin\theta\cos\theta = 2\sin(2\theta) Thus, the integral becomes: 02x(4x)dx=0π42sin(2θ)4sin(2θ)dθ=80π4sin2(2θ)dθ\int_0^2 \sqrt{x(4-x)} \, dx = \int_0^{\frac{\pi}{4}} 2\sin(2\theta) \cdot 4\sin(2\theta) \, d\theta = 8 \int_0^{\frac{\pi}{4}} \sin^2(2\theta) \, d\theta

Step 3: Solve the integral

We can use the identity sin2(2θ)=1cos(4θ)2\sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2} to simplify: 0π4sin2(2θ)dθ=120π4(1cos(4θ))dθ\int_0^{\frac{\pi}{4}} \sin^2(2\theta) \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{4}} (1 - \cos(4\theta)) \, d\theta This gives: [ \frac{1}{2} \left[ \theta - \frac{\sin(4\theta)}{4} \right]_0^{\frac{\pi}{4}} = \frac{1}{2} \left[ \frac{\pi}{4} - 0 \right] = \frac{\pi}{8} ]

Step 4: Final answer

Multiplying everything together, we get: 8π8=π8 \cdot \frac{\pi}{8} = \pi

Thus, the exact value of the integral is π\boxed{\pi}.

Would you like any more details, or have any questions about this solution?

Here are 5 related questions:

  1. How do trigonometric identities simplify integrals involving squares of sine or cosine functions?
  2. What other types of substitutions are useful for integrals of this kind?
  3. Can this method be used for other limits or different forms of the integrand?
  4. How does the double-angle identity help simplify integrals?
  5. How would the integral change if the limits of integration were different?

Tip: Always remember to change the limits of integration when switching variables!

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Math Problem Analysis

Mathematical Concepts

Trigonometric Substitution
Integral Calculus
Trigonometric Identities

Formulas

Trigonometric substitution: \(x = 4\sin^2\theta\)
Double angle identity: \(\sin(2\theta) = 2\sin\theta\cos\theta\)
Simplified integrand: \(\sqrt{16\sin^2\theta\cos^2\theta} = 4\sin\theta\cos\theta\)

Theorems

Double-Angle Identity
Integration by Substitution

Suitable Grade Level

Grades 11-12, College Level