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Magazine Chapter 8: Problem 7
Evaluate the integrals in Exercises \(1-28\). $$ \int \sqrt{25-t^{2}} d t $$
Short Answer
Expert verified
\( \int \sqrt{25-t^2} \, dt = \frac{25}{2}\arcsin\left(\frac{t}{5}\right) + \frac{t}{2}\sqrt{25-t^2} + C \).
Step by step solution
01
Substitution
To solve the integral \( \int \sqrt{25-t^2} \, dt \), we use a trigonometric substitution. Let \( t = 5 \sin \theta \). Then, \( dt = 5 \cos \theta \, d\theta \). Substitute these into the integral.
02
Simplify the Integrand
Substitute to get \( \int \sqrt{25-(5\sin\theta)^2} \cdot 5\cos\theta \, d\theta \). This simplifies to \( 25\cos\theta \), and the integrand becomes \( \int 25\cos^2\theta \, d\theta \).
03
Use Trigonometric Identity
Recall the identity \( \cos^2 \theta = \frac{1+\cos 2\theta}{2} \). Substitute to obtain \( \int 25 \left( \frac{1+\cos 2\theta}{2} \right) \, d\theta = \int \frac{25}{2} \, d\theta + \int \frac{25}{2} \cos 2\theta \, d\theta \).
04
Integrate the Terms
The first integral, \( \int \frac{25}{2} \, d\theta \), gives \( \frac{25}{2}\theta \). The second integral, \( \int \frac{25}{2} \cos 2\theta \, d\theta \), gives \( \frac{25}{4} \sin 2\theta \) using the substitution \( u = 2\theta \). Combine the results: \( \frac{25}{2}\theta + \frac{25}{4}\sin 2\theta \).
05
Back-Substitute \( \theta \) to \( t \)
We originally set \( t = 5 \sin \theta \), so \( \sin \theta = \frac{t}{5} \). The angle \( \theta \) can be expressed as \( \theta = \arcsin(\frac{t}{5}) \). Also, \( \sin 2\theta = 2\sin\theta\cos\theta \), and \( \cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-(\frac{t}{5})^2} = \frac{\sqrt{25-t^2}}{5} \). Substitute back to get the integral in terms of \( t \): \( \frac{25}{2}\arcsin\left(\frac{t}{5}\right) + \frac{t}{2}\sqrt{25-t^2} + C \), where \( C \) is the integration constant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Substitution
Trigonometric substitution is a key integration technique often employed when dealing with integrals involving square roots of quadratic expressions, like \( \sqrt{25-t^2} \). In our problem, we make the substitution \( t = 5 \sin \theta \). Why do we use trigonometric substitution here? This is because the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) can simplify the integrand when rewritten appropriately.
- By setting \( t = 5 \sin \theta \), we derive \( dt = 5 \cos \theta \, d\theta \).
- The square root term \( \sqrt{25-t^2} \) then becomes \( \sqrt{25 - (5\sin \theta)^2} = 5 \cos \theta \).
- This substitution transforms our integral into \( \int 25\cos^2 \theta \, d\theta \), which is much easier to integrate compared to its original form.
Definite and Indefinite Integrals
The process of integration can result in two types: definite and indefinite integrals. An indefinite integral, as seen in this exercise, represents an antiderivative, adding a constant \( C \) because the original integral can represent a whole family of functions.
- In our problem, the indefinite integral of our transformed function \( \int 25\cos^2 \theta \, d\theta \) first leads us step by step to results like \( \frac{25}{2}\theta + \frac{25}{4}\sin 2\theta + C \).
- This means we find a general form \( g(\theta) \), which on differentiating will return the expression under the integral.
- The addition of \( C \) accounts for all vertical shifts in possible antiderivative functions, because derivatives of constants are zero.
Trigonometric Identities
Understanding trigonometric identities is essential when integrating trigonometric functions. Among these, \( \cos^2 \theta = \frac{1+\cos 2\theta}{2} \) plays a fundamental role in simplifying integrals like the one in this exercise.
- To integrate \( \cos^2 \theta \), we apply this identity, splitting it as \( \int \frac{1+\cos 2\theta}{2} \, d\theta = \int \frac{1}{2} \, d\theta + \int \frac{\cos 2\theta}{2} \, d\theta \).
- This separation uses the identity to turn a difficult integral into a sum of simpler ones that we can compute more directly.
- These simpler forms help us integrate using direct formulas: \( \int \cos(ax) \, dx \) leads to a sine function while \( \int 1 \, dx \) gives a linear function, both of which are straightforward solutions.
