Problem 18 - Essential Calculus

Question

Find the integral. $$ \int x \sqrt{16-4 x^{2}} d x $$

Step-by-Step Solution

Answer

The integral of \(x \sqrt{16-4x^{2}}\) dx is \(-8x \sqrt{16-x^2} + \frac{8}{3}(16-x^2)^{\frac{3}{2}} + C\) where C is the constant of integration.

Step 1: Identify the Correct Substitution

The denominator part of the integrand under the square root \(\sqrt{16-4x^2}\) resembles the formula for Pythagorean identity in trigonometry \(\sin^2\theta + \cos^2\theta = 1\). For easy handling of the integral, one introduces a substitution \(2x=4\sin\theta\) or \(x=2\sin\theta\). The differential dx can be computed as \(dx=2\cos\theta d\theta\).

Step 2: Substitute and Simplify the Integral

Let's substitute \(x=2\sin\theta\) and \(dx=2\cos\theta d\theta\) into the integral \[ \int x \sqrt{16-4x^2} dx = \int 2\sin\theta\sqrt{16-16\sin^2\theta} * 2\cos\theta d\theta.\] After simplifying, we get \[ \int 2\sin\theta*(4|\cos\theta|)*2\cos\theta d\theta = 8\int\sin\theta \cos^2\theta d\theta.\] Here, \( |\cos\theta|\) is taken as \(\cos\theta\) since the restriction that we have on \(\theta\), after substitution is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) which means \(\cos \theta >= 0 \) avoiding the need to split the integral.

Step 3: Use Trigonometric Identity

Next, we can express \(\cos^2\theta\) in terms of \(\sin\theta\) using the Pythagorean identity \(\cos^2\theta = 1 - \sin^2\theta\), and the integral becomes\[ \int 8\sin\theta (1-\sin^2\theta) d\theta = 8 \int (\sin\theta-\sin^3\theta) d\theta.\]

Step 4: Evaluate the Integral

The integral can now be integrated directly, yielding\[ 8 \int (\sin\theta - \sin^3\theta) d\theta = 8 (-cos\theta + \frac{cos^3\theta}{3}) + C, \]where \(C\) is the constant of integration.

Step 5: Back-Substitute to Original Variable

Now we should express our answer in terms of the original variable \(x\). Remembering that we used the substitution \(x=2\sin\theta\), to find \(\cos\theta\), we use \(\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-\frac{x^2}{4}}\), we substitute back and simplify, finally we reach that the integral is equal to\[ -8x \sqrt{16-x^2} + \frac{8}{3}(16-x^2)^{\frac{3}{2}} + C.\]