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Chapter 6: Problem 35

Evaluate the integral using (a) the given integration limits and (b) the limits obtained by trigonometric substitution. $$ \int_{0}^{3} \frac{x^{3}}{\sqrt{x^{2}+9}} d x $$

Short Answer

Expert verified
The integral can be solved using trigonometric substitution which simplifies the integral. However, the integral needs to be evaluated with the original limits as well to see if the result differs from the one obtained by substitution.

Step by step solution

01

Identifying the form of the integral

Typically, for integrals which have the form \(\sqrt{a^2+x^2}\) or \(\sqrt{x^2+a^2}\), setting \(x = a\tan{\theta}\) is beneficial. Here, the constant \(a^2\) is 9, thus, \(a = 3\). So, we can let \(x = 3\tan{\theta}\). Consequently, \(dx = 3\sec^2{\theta}d\theta\).
02

Changing the limits

The original limits of the integral are x = 0 to x = 3. We need to change these x-values to the corresponding 胃-values. When \(x = 0\), \(\theta = \arctan{(0/3)} = 0\). When \(x = 3\), \(\theta = \arctan{(3/3)} = \arctan{1} = \pi/4\). So, the new limits are \(\theta = 0\) to \(\theta = \pi/4\).
03

Substituting x, dx, and the limits into the integral

Substitute \(x=3\tan{\theta}\) and \(dx = 3\sec^2{\theta}d\theta\), and the new limits into the integral and simplify.
04

Evaluating the integral

To solve this integral, we may use integrals of powers of secant, and definite integral properties.
05

Evaluating integral with original limits

Evaluate the integral with the original limits provided without change. This could be achieved using a standard calculus integral table or software.

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