Math Problem Statement

integral of square root 16-x^2

Solution

The integral you're looking to solve is:

16x2dx\int \sqrt{16 - x^2} \, dx

This is a standard integral that can be solved using a trigonometric substitution. Here's the step-by-step approach:

Step 1: Use Trigonometric Substitution

We will use the substitution x=4sin(θ)x = 4 \sin(\theta), since 16x2\sqrt{16 - x^2} resembles the identity 1sin2(θ)=cos(θ)\sqrt{1 - \sin^2(\theta)} = \cos(\theta).

Thus, let: x=4sin(θ)x = 4 \sin(\theta) Then, dx=4cos(θ)dθdx = 4 \cos(\theta) \, d\theta

Step 2: Simplify the Integrand

Substitute x=4sin(θ)x = 4 \sin(\theta) into the integrand:

16x2=1616sin2(θ)=4cos(θ)\sqrt{16 - x^2} = \sqrt{16 - 16\sin^2(\theta)} = 4\cos(\theta)

So the integral becomes:

16x2dx=4cos(θ)4cos(θ)dθ=16cos2(θ)dθ\int \sqrt{16 - x^2} \, dx = \int 4\cos(\theta) \cdot 4\cos(\theta) \, d\theta = \int 16\cos^2(\theta) \, d\theta

Step 3: Use a Trigonometric Identity

We can simplify cos2(θ)\cos^2(\theta) using the half-angle identity:

cos2(θ)=1+cos(2θ)2\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}

So, the integral becomes:

16cos2(θ)dθ=161+cos(2θ)2dθ=8(1+cos(2θ))dθ\int 16\cos^2(\theta) \, d\theta = \int 16 \cdot \frac{1 + \cos(2\theta)}{2} \, d\theta = 8 \int (1 + \cos(2\theta)) \, d\theta

Step 4: Integrate

Now, integrate:

8(1+cos(2θ))dθ=8[θ+sin(2θ)2]+C8 \int (1 + \cos(2\theta)) \, d\theta = 8 \left[ \theta + \frac{\sin(2\theta)}{2} \right] + C

Step 5: Reverse the Substitution

Recall that x=4sin(θ)x = 4 \sin(\theta), so sin(θ)=x4\sin(\theta) = \frac{x}{4}. Now, solve for θ\theta using θ=arcsin(x4)\theta = \arcsin\left(\frac{x}{4}\right).

Also, sin(2θ)=2sin(θ)cos(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta), and from the Pythagorean identity cos(θ)=1sin2(θ)=1(x4)2=16x24\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{x}{4}\right)^2} = \frac{\sqrt{16 - x^2}}{4}.

Thus, sin(2θ)=2x416x24=x16x28\sin(2\theta) = 2 \cdot \frac{x}{4} \cdot \frac{\sqrt{16 - x^2}}{4} = \frac{x \sqrt{16 - x^2}}{8}.

Now substitute these into the integral result:

[ 8 \left[ \theta + \frac{\sin(2\theta)}{2} \right] = 8 \left[ \arcsin\left( \frac{x}{4} \right) + \frac{x \sqrt{16 - x^2}}{16} \right] + C ]

Final Answer:

16x2dx=8arcsin(x4)+x16x22+C\int \sqrt{16 - x^2} \, dx = 8 \arcsin\left( \frac{x}{4} \right) + \frac{x \sqrt{16 - x^2}}{2} + C

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


Related Questions:

  1. How does the trigonometric substitution method help simplify integrals involving square roots of quadratic expressions?
  2. Can you derive the integral of a2x2\sqrt{a^2 - x^2} using a similar substitution method?
  3. What is the significance of the half-angle identity in reducing trigonometric integrals?
  4. How do you handle definite integrals with square roots of quadratic functions using trigonometric substitution?
  5. Can the result be applied to other integrals, like 9x2dx\int \sqrt{9 - x^2} \, dx?

Tip: Always remember that recognizing a standard form like a2x2\sqrt{a^2 - x^2} can greatly simplify the process of integration. Trigonometric substitution is a powerful tool for this!

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Math Problem Analysis

Mathematical Concepts

Integration
Trigonometric Substitution
Trigonometric Identities

Formulas

cos^2(θ) = (1 + cos(2θ))/2
sin^2(θ) + cos^2(θ) = 1
Integration formula for √(a^2 - x^2): ∫√(a^2 - x^2) dx = (a^2/2)arcsin(x/a) + (x/2)√(a^2 - x^2) + C

Theorems

Pythagorean Identity
Trigonometric Substitution Theorem

Suitable Grade Level

Undergraduate Calculus (or advanced high school calculus)