Problem 17 - Essential Calculus

Question

Find the integral. $$ \int \frac{1}{\sqrt{16-x^{2}}} d x $$

Step-by-Step Solution

Answer

\(\arcsin\left(\frac{x}{4}\right) + C\)

Step 1: Trigonometric substitution

Start with the function \(\int \frac{1}{\sqrt{16-x^{2}}} dx\). In this case, \(a = 4\) because the square root term is in terms of \(16 = 4^2\). So, we make a trigonometric substitution of \(x = 4\sin(\theta)\). Then, differentiate \(x = 4\sin(\theta)\) with respect to \(\theta\) to get \(dx = 4\cos(\theta) d\theta\). Now, replace \(x\) with \(4\sin(\theta)\) in the function and \(dx\) with \(4\cos(\theta) d\theta\). The function will become \(\int \frac{1}{\sqrt{16-(4\sin(\theta))^2}} \cdot 4\cos(\theta) d\theta\).

Step 2: Simplify and integrate

Simplify the function under the square root to get \(\int \frac{1}{\sqrt{16(1-\sin^2(\theta))}} \cdot 4\cos(\theta) d\theta\). Further simplification due to \(1-\sin^2(\theta) = \cos^2(\theta)\) gives \(\int \frac{4\cos(\theta)}{\sqrt{16\cos^2(\theta)}} d\theta\), which further simplifies to \(\int d\theta\). The integral of \(d\theta\) is \(\theta\) + C.

Step 3: Substitute \(\theta\) back in terms of \(x\)

Now replace \(\theta\) with \(\arcsin\left(\frac{x}{4}\right)\) because from the original substitution \(x = 4\sin(\theta)\), \(\theta = \arcsin\left(\frac{x}{4}\right)\). Thus, the final answer is \(\arcsin\left(\frac{x}{4}\right) + C\).