Problem 3
Question
$$ \int \frac{d x}{\sqrt{a^{2}-b^{2} x^{2}}}=\frac{1}{b} \sin ^{-1} \frac{b x}{a} $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \frac{1}{b} \sin^{-1} \frac{bx}{a} + C. \)
1Step 1: Evaluate the Type of Integral
This integral is in the form of \[ \int \frac{dx}{\sqrt{a^2 - b^2x^2}} \] which is a standard form of an integral involving the inverse sine function. Recognizing this form helps identify the appropriate method and solution.
2Step 2: Recall the Standard Integral Formula
The standard integral formula for \[ \int \frac{dx}{\sqrt{a^2 - x^2}} \] is \[ \sin^{-1} \frac{x}{a} + C. \] Since our integral involves \(b^2x^2\), a substitution is needed to match this form.
3Step 3: Perform Substitution
Let \(u = bx\). Then, \(du = bdx\) or \(dx = \frac{du}{b}\). Substitute into the integral to get \[ \int \frac{1}{b} \frac{du}{\sqrt{a^2 - u^2}}. \]
4Step 4: Integrate with Respect to 'u'
The integral now matches the standard form: \[ \int \frac{du}{\sqrt{a^2 - u^2}} \] which is \[ \sin^{-1} \frac{u}{a} + C. \] Hence, we have \( \frac{1}{b} \sin^{-1} \frac{u}{a} + C \).
5Step 5: Substitute Back to Original Variable
Replace \(u\) with \(bx\) to get the original variables back: \[ \frac{1}{b} \sin^{-1} \frac{bx}{a} + C. \] This matches the given solution form.
Key Concepts
Inverse Trigonometric FunctionsDefinite and Indefinite IntegralsTrigonometric Substitution
Inverse Trigonometric Functions
Inverse trigonometric functions, such as the inverse sine function often represent angles in trigonometry that correspond to specific values of sine functions. In calculus, these functions become essential when managing particular kinds of integrals.
For example, \(\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + C\)\ showcases the use of the inverse sine function to solve the integral, providing the antiderivative for expressions structured similarly. In the given exercise, the expression used is slightly different due to the additional factor \(b^2x^2\), but still involves transforming into a form that uses the inverse sine function.
This connection helps bridge the gap between trigonometric concepts and calculus, allowing us to manage more complex integrals by relating them to known functions.
For example, \(\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + C\)\ showcases the use of the inverse sine function to solve the integral, providing the antiderivative for expressions structured similarly. In the given exercise, the expression used is slightly different due to the additional factor \(b^2x^2\), but still involves transforming into a form that uses the inverse sine function.
This connection helps bridge the gap between trigonometric concepts and calculus, allowing us to manage more complex integrals by relating them to known functions.
Definite and Indefinite Integrals
Integrals come in two main types: definite and indefinite. An indefinite integral represents the collection of all antiderivatives of a function and is accompanied by a constant of integration, symbolized as \(C\).
Indefinite integrals appear without specific limits of integration, giving us a general solution: \(\int f(x) \, dx = F(x) + C\).\ This shows the general formula without boundaries. On the other hand, definite integrals have bounds and provide the net area under a curve between these two limits.
Indefinite integrals appear without specific limits of integration, giving us a general solution: \(\int f(x) \, dx = F(x) + C\).\ This shows the general formula without boundaries. On the other hand, definite integrals have bounds and provide the net area under a curve between these two limits.
- Indefinite Integral: Represents a family of functions. Always added with a constant \(C\).
- Definite Integral: Evaluates the area between two points, resulting in a specific numerical value.
Trigonometric Substitution
Trigonometric substitution is a powerful technique used in calculus to simplify integrals by substituting a trigonometric function for a variable. This approach is particularly effective when dealing with integrals involving radicals of the form \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\).
In the exercise we examined, trigonometric substitution played a key role in simplifying the integral. The substitution \(u = bx\) was made to transform the integral into a standard form that can be easily solved using the inverse sine function.
Here's a simplified breakdown of the substitution steps:
In the exercise we examined, trigonometric substitution played a key role in simplifying the integral. The substitution \(u = bx\) was made to transform the integral into a standard form that can be easily solved using the inverse sine function.
Here's a simplified breakdown of the substitution steps:
- Choose a substitution that will simplify the given integral.
- Convert the entire integral in terms of the new variable.
- Simplify the integral into a known form using inverse trigonometric identities or formulae.
- Integrate and then revert the substitution to express the answer in the terms of the original variable.
Other exercises in this chapter
Problem 1
Fvaluate \(\int \frac{e^{x}\left(x^{2}+1\right) d x}{(x+1)^{2}}\)
View solution Problem 2
Fvaluate \(\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x\)
View solution Problem 3
$$ \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right| \text { or } \frac{1}{a} \tan h^{-1}\left(\frac{x}{a}\right) $$
View solution Problem 3
Lvaluate \(\int \sqrt{\frac{a+x}{x}} d x\)
View solution