Chapter 8: Problem 6
Show that $$ \int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}}=\frac{1}{a} \sec ^{-1}\left(\frac{u}{a}\right)+C $$
Short Answer
Expert verified
\( \int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}} = \frac{1}{a} \sec^{-1}\frac{u}{a} + C \)
Step by step solution
01
Recognize the Integral Form
The given integral \( \int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}} \) suggests that a trigonometric substitution may simplify it. Notice that \( u^{2} - a^{2} \) in the denominator resembles forms where trigonometric identities can be applied.
02
Apply Trigonometric Substitution
Let \( u = a \sec \theta \). Then we have \( \mathrm{d} u = a \sec \theta \tan \theta \mathrm{d} \theta \). Substituting these values into the integral will transform it into a trigonometric integral.
03
Simplify the Integral
Substitute \( u = a \sec \theta \) and \( \mathrm{d} u = a \sec \theta \tan \theta \mathrm{d} \theta \) into the integral: \[ \int \frac{a \sec \theta \tan \theta \mathrm{d} \theta}{a \sec \theta \sqrt{a^{2} \sec^{2}\theta - a^{2}}} \].
04
Simplify the Denominator
Simplify the square root in the denominator: \[ \sqrt{a^{2} \sec^{2} \theta - a^{2}} = a \sqrt{\frac{\sec^{2}\theta-1}{\sec^{2}\theta}} = a \tan \theta \]. This simplifies our integral to: \[ \int \frac{a \sec \theta \tan \theta \mathrm{d} \theta}{a \sec \theta \tan \theta} \].
05
Integrate
After canceling common terms we get: \[ \int \frac{a \sec \theta \tan \theta \mathrm{d} \theta}{a \sec \theta \tan \theta} = \int \mathrm{d} \theta = \theta + C \].
06
Back Substitute
Recall our substitution \( u = a \sec \theta \). Solving for \theta, we get \( \theta = \sec^{-1}\frac{u}{a} \). Substitute back to find: \ \[ \theta + C = \sec^{-1}\frac{u}{a} + C \].
07
Finalize the Solution
Therefore, we have shown that: \[ \int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}} = \frac{1}{a} \sec^{-1}\frac{u}{a} + C. \]
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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 technique used in calculus to simplify integrals. It involves replacing a variable with a trigonometric function to use trigonometric identities for simplification. This method is particularly useful for integrals involving square roots of quadratic expressions, such as the form \(\sqrt{u^{2}-a^{2}}\).
In the example provided, the integral to simplify is \(\int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}}\). Here, we recognize that by using the trigonometric identity \(\sec^2 \theta - 1 = \tan^2 \theta\), we can choose substitution to simplify the integral.
In the example provided, the integral to simplify is \(\int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}}\). Here, we recognize that by using the trigonometric identity \(\sec^2 \theta - 1 = \tan^2 \theta\), we can choose substitution to simplify the integral.
- Set \(u = a \sec \theta\)
- So, \(\mathrm{d} u = a \sec \theta \tan \theta \mathrm{d} \theta\)
definite and indefinite integrals
An integral can be categorized as either definite or indefinite:
The integral given in the exercise, \(\int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}}\), is an indefinite integral. Solving it involves finding the antiderivative of the function inside the integral and including a constant of integration \(C\). Through substitution and simplification, we ultimately obtain the indefinite form \( \frac{1}{a} \sec ^{-1}\left(\frac{u}{a}\right) + C \).
- Indefinite integrals represent a family of functions and include a constant of integration \(C\). It looks like \(\int f(x) \, \mathrm{d} x = F(x) + C\), where \(F(x)\) is the antiderivative of \(f(x)\).
- Definite integrals, on the other hand, have upper and lower limits. They give us a real number representing the area under the curve between those limits. It’s denoted as \(\int_{a}^{b} f(x) \, \mathrm{d} x = F(b) - F(a)\).
The integral given in the exercise, \(\int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}}\), is an indefinite integral. Solving it involves finding the antiderivative of the function inside the integral and including a constant of integration \(C\). Through substitution and simplification, we ultimately obtain the indefinite form \( \frac{1}{a} \sec ^{-1}\left(\frac{u}{a}\right) + C \).
calculus techniques
Several calculus techniques are employed while solving integrals, and they considerably streamline the process:
- Substitution: This technique, also known as u-substitution, simplifies an integral by changing the variable of integration.
- Parts: Integration by parts is derived from the product rule of differentiation and helps in integrating products of functions.
- Partial Fractions: This technique is used to break down complex rational functions into simpler fractions that can be easily integrated.
- Substituting \(u = a \sec \theta\) transforms the integral to a simpler form.
- Simplifying the trigonometric expressions to make integration feasible.
- Reverting back to the original variable using back substitution.