Problem 16 - Essential Calculus

Question

Find the integral. $$ \int \frac{x}{\sqrt{9-x^{2}}} d x $$

Step-by-Step Solution

Answer

The integral of the given function, \( \int \frac{x}{\sqrt{9-x^2}} dx\) , is \(-\frac{1}{4} + \frac{x^2}{4} + C\).

Step 1: Initiate Substitution

We need to use trigonometric substitution. Let \(x = 3\sin \theta\). From the Pythagorean identity \(1 = \sin^2\theta + \cos^2\theta\), we have \(9-x^2 = 9\cos^2\theta\), then \(\sqrt{9-x^2} = 3\cos \theta\).

Step 2: Calculate Derivative

Next, find the derivative of \(x = 3\sin \theta\) which is \(dx = 3\cos \theta d\theta\).

Step 3: Replace dx in Integral

Substitute \(x = 3\sin \theta\) and \(dx = 3\cos \theta d\theta\) into the integral.

Step 4: Simplify the Integral

The integral becomes \(\int \sin \theta \cdot 3\cos \theta d\theta\). With the double angle identity \(\sin 2\theta = 2\sin\theta \cos\theta\), it can be rewritten as \(\frac{1}{2}\int \sin 2\theta d\theta\).

Step 5: Solve the Integral

The integral \(\frac{1}{2}\int \sin 2\theta d\theta\) simplifies to \(-\frac{1}{4}\cos 2\theta + C\).

Step 6: Convert back to x Terms

Change the term back to \(x\) using the double angle identity \(\cos2\theta = 1- 2\sin^2\theta\). This will result in \(-\frac{1}{4} + \frac{x^2}{4} + C\).