Problem 29
Question
Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{2 s d s}{\sqrt{1-s^{4}}} $$
Step-by-Step Solution
Verified Answer
\( \arcsin(s^2) + C \)
1Step 1: Choose the substitution
To solve the integral \( \int \frac{2s \, ds}{\sqrt{1-s^4}} \), we will choose a substitution to simplify the expression under the square root. Let's use the substitution \( u = s^2 \). This gives us \( du = 2s \, ds \), which matches the numerator.
2Step 2: Rewrite the integral
Substitute \( u = s^2 \) into the integral: \( s^2 = u \) and \( 2s \, ds = du \), so the integral becomes \( \int \frac{du}{\sqrt{1-u^2}} \).
3Step 3: Identify the substitution's result
The integral \( \int \frac{du}{\sqrt{1-u^2}} \) is a standard integral that corresponds to \( \arcsin(u) + C \) where \( C \) is the constant of integration.
4Step 4: Back-substitute to original variable
Since we used the substitution \( u = s^2 \), we substitute back to get \( \arcsin(s^2) + C \) as the integral evaluated in terms of the original variable \( s \).
Key Concepts
Substitution MethodDefinite IntegralsInverse Trigonometric Functions
Substitution Method
The substitution method is a powerful tool in calculus, especially for integration. It involves changing variables in an integral, which makes it easier to evaluate. Essentially, you're looking for a substitution that simplifies the challenging part of the integral.
In our example, we want to solve \( \int \frac{2s \, ds}{\sqrt{1-s^4}} \). At first glance, this might seem complicated due to the square root of a polynomial in the denominator. To simplify, we use a substitution: \( u = s^2 \).
This choice helps for several reasons:
Ultimately, the substitution transforms a difficult integral into an easier, more familiar one.
In our example, we want to solve \( \int \frac{2s \, ds}{\sqrt{1-s^4}} \). At first glance, this might seem complicated due to the square root of a polynomial in the denominator. To simplify, we use a substitution: \( u = s^2 \).
This choice helps for several reasons:
- By taking the derivative, we get \( du = 2s \, ds \), which simplifies the numerator.
- Substituting \( u = s^2 \) into the denominator makes it a form we recognize.
Ultimately, the substitution transforms a difficult integral into an easier, more familiar one.
Definite Integrals
Definite integrals give the net area under a curve between two points, and they evaluate to a number rather than a function. However, the integral in our example is indefinite, as it doesn't have specified limits. Instead, it represents a family of functions, all identical except for a constant addition.
While our case doesn't deal directly with definite integrals, the computation process is similar. After finding an antiderivative, definite integrals require evaluation at specified bounds. This would involve substituting those limits into the antiderivative, finding the difference, and providing the net area
Understanding indefinite integrals, like the example \( \arcsin(s^2) + C \), helps with definite integrals. It prepares you to tackle those problems by teaching you how to find antiderivatives.
While our case doesn't deal directly with definite integrals, the computation process is similar. After finding an antiderivative, definite integrals require evaluation at specified bounds. This would involve substituting those limits into the antiderivative, finding the difference, and providing the net area
Understanding indefinite integrals, like the example \( \arcsin(s^2) + C \), helps with definite integrals. It prepares you to tackle those problems by teaching you how to find antiderivatives.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \( \arcsin \), \( \arccos \), and \( \arctan \), are critical in calculus, particularly in integration. They reverse trigonometric functions, calculating the angle (or input) when given the output of a trigonometric function.
In the solution to our exercise, the integral \( \int \frac{du}{\sqrt{1-u^2}} \) corresponds to \( \arcsin(u) + C \). This is because \( \arcsin \) represents the antiderivative of that form. Recognizing these forms makes it easier to integrate, as you'd match similar expressions to inverses.
These functions often appear when dealing with expressions under square roots or when evaluating integrals with specific forms. By understanding inverse trigonometric functions, you can easily integrate expressions that seem complex at first glance.
In the solution to our exercise, the integral \( \int \frac{du}{\sqrt{1-u^2}} \) corresponds to \( \arcsin(u) + C \). This is because \( \arcsin \) represents the antiderivative of that form. Recognizing these forms makes it easier to integrate, as you'd match similar expressions to inverses.
These functions often appear when dealing with expressions under square roots or when evaluating integrals with specific forms. By understanding inverse trigonometric functions, you can easily integrate expressions that seem complex at first glance.
Other exercises in this chapter
Problem 29
Evaluate the integrals by using a substitution prior to integration by parts. \(\int \sin (\ln x) d x\)
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In Exercises \(29-34,\) perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral. $$ \int
View solution Problem 30
In Exercises \(29-36\) , use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$ \int_{\ln (3 / 4)}^{\ln (4 / 3)} \f
View solution Problem 30
Use the table of integrals at the back of the book to evaluate the integrals. \(\int \frac{\sqrt{3 t+9}}{t} d t\)
View solution