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Chapter 8: Problem 8

Evaluate the integrals in Exercises \(1-14\) $$ \int \sqrt{1-9 t^{2}} d t $$

Short Answer

Expert verified
\( \frac{1}{6} ( \arcsin(3t) + 3t\sqrt{1-9t^2} ) + C \)

Step by step solution

01

Choose an Appropriate Substitution

Notice the integral \( \int \sqrt{1-9t^2} \, dt \) resembles the form of \( \sqrt{a^2 - x^2} \), suggesting a trigonometric substitution. Let \( t = \frac{1}{3} \sin \theta \). Thus, \( dt = \frac{1}{3} \cos \theta \, d\theta \).
02

Substitute and Simplify

Substituting into the integral, we get:\[\int \sqrt{1-9\left(\frac{1}{3}\sin\theta\right)^2} \cdot \frac{1}{3}\cos\theta \, d\theta = \int \sqrt{1-\sin^2\theta} \cdot \frac{1}{3}\cos\theta \, d\theta.\]This simplifies to:\[\int \cos\theta \cdot \frac{1}{3}\cos\theta \, d\theta = \frac{1}{3} \int \cos^2 \theta \, d\theta.\]
03

Use Trigonometric Identities

Apply the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \) to the integral:\[\frac{1}{3} \int \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{1}{6} \int (1 + \cos 2\theta) \, d\theta.\]
04

Integrate

Separate the integral:\[\frac{1}{6} \left( \int 1 \, d\theta + \int \cos 2\theta \, d\theta \right) = \frac{1}{6} \left( \theta + \frac{1}{2} \sin 2\theta + C \right).\]
05

Substitute Back

We need to revert the substitution \( t = \frac{1}{3} \sin \theta \). Therefore, \( \sin \theta = 3t \) and \( \theta = \arcsin(3t) \). Also, \( \sin 2\theta = 2\sin \theta \cos \theta = 2(3t)\sqrt{1-(3t)^2} = 6t\sqrt{1-9t^2} \). Thus, the integral becomes:\[\frac{1}{6} \left( \arcsin(3t) + 3t\sqrt{1-9t^2} + C \right).\]
06

Final Answer

Combining these results, we have:\[\frac{1}{6} \left( \arcsin(3t) + 3t\sqrt{1-9t^2} \right) + C.\] This is the evaluated integral for \( \int \sqrt{1-9t^2} \, dt \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Integration techniques are mathematical methods used to evaluate complex integrals. In this particular case, we're dealing with a form \( \int \sqrt{1-9t^2} \, dt \) which resembles \( \sqrt{a^2 - x^2} \). Trigonometric substitution is an effective technique for handling such integrals.

Key Steps in Trigonometric Substitution:
  • Identify the form: Recognize that \( \sqrt{1-x^2} \) can be approached with \( x = \sin \theta \), which simplifies \( \sqrt{1-x^2} = \cos \theta \).
  • Substitute: Convert variables, making substitutions for both the variable and differential (e.g., \( t = \frac{1}{3} \sin \theta \)).
  • Simplify: Use trigonometric identities to simplify the expression.

Trigonometric substitution converts a tough integral into a more manageable one. It often simplifies the process of integrating functions involving square roots of quadratic expressions.
Trigonometric Identities
Trigonometric identities play a crucial role in simplifying integrals, especially when using trigonometric substitution.

In our example, we need to handle \( \cos^2 \theta \), which arises after the trigonometric substitution.
To simplify this, we use the identity:
  • Cosine squared identity: \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \)
This identity is essential for breaking down the integral into simpler parts, making it possible to integrate term by term.
By converting \( \cos^2 \theta \) into \( \frac{1 + \cos 2\theta}{2} \), the integral splits into two straightforward integrals, \( \int 1 \, d\theta \) and \( \int \cos 2\theta \, d\theta \).
Trigonometric identities are fundamental tools in integration, helping us navigate complex trigonometric functions and simplify them for easy evaluation.
Inverse Trigonometric Functions
Inverse trigonometric functions can appear as final results or intermediate steps when solving integrals involving trigonometric substitution.

In the final stage of solving our integral, it's crucial to substitute back to the original variable. The relation originally defined gives us \( \theta = \arcsin(3t) \).
Substitution Back to Original Variable:
  • After using the identity \( \sin \theta = 3t \), we find \( \theta \) as \( \arcsin(3t) \).
  • This result, \( \arcsin(3t) \), is incorporated back into the expression to reflect the original variable \( t \).
Inverse trigonometric functions like \( \arcsin \) help us express angles in terms of original variables, such as \( t \).
These functions ensure that the final result remains relevant to the problem's initial setup, providing a complete solution.

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Most popular questions from this chapter

In Exercises \(27-40\) , use a substitution to change the integral into one you can find in the table. Then evaluate the integral. $$ \int \frac{\sqrt{2-x}}{\sqrt{x}} d x $$

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