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College Algebra

10th Edition

ISBN:9781337282291

Author:Ron Larson

Publisher:Ron Larson

ChapterP: Prerequisites

SectionP.3: Polynomials And Special Products

Problem 5ECP

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Question

Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

Expert Solution

Step 1: Given

$\int_{0}^{2} \frac{x^{2}}{x^{2} + 4} d x$ .

Step 2: To determine

Let, $\int_{0}^{2} \frac{x^{2}}{x^{2} + 4} d x$

Put $x = 2 \tan θ \Rightarrow d x = 2 s e c^{2} θ d θ$

When $x = 0 then θ = 0$ and when $x = 2 then θ = \frac{π}{4}$ .

$\begin{array}{rcl} \int_{0}^{2} \frac{x^{2}}{x^{2} + 4} d x & = & \int_{0}^{\frac{π}{4}} \frac{{(2 \tan θ)}^{2}}{{(2 \tan θ)}^{2} + 4} d x \\ = & \int_{0}^{\frac{π}{4}} \frac{4 \tan^{2} θ}{4 \tan^{2} θ + 4} 2 s e c^{2} θ d θ \\ = & \int_{0}^{\frac{π}{4}} \frac{4 \tan^{2} θ}{4 (\tan^{2} θ + 1)} 2 s e c^{2} θ d θ \\ = & 2 \int_{0}^{\frac{π}{4}} \tan^{2} θ d θ \\ = & 2 {[- θ + \tan θ]}_{0}^{\frac{π}{4}} \\ = & 2 [- \frac{π}{4} + \tan (\frac{π}{4}) + 0 - \tan (0)] \\ = & 2 [- \frac{π}{4} + 1] \end{array}$

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