Problem 7
Question
Evaluate the integrals. \(\int \tan ^{-1} y d y\)
Step-by-Step Solution
Verified Answer
\( y \tan^{-1}y - \frac{1}{2} \ln |1+y^2| + C \) is the result.
1Step 1: Understand the Problem
We need to evaluate the integral \( \int \tan^{-1} y \; dy \). This requires finding the antiderivative of the inverse tangent function with respect to \( y \).
2Step 2: Integration by Parts Setup
Integration by parts is a useful technique here. It is defined as \( \int u \; dv = uv - \int v \; du \). We choose \( u = \tan^{-1}y \) and \( dv = dy \).
3Step 3: Differentiate and Integrate Parts
Differentiate \( u = \tan^{-1}y \) to find \( du = \frac{1}{1+y^2} dy \). Integrate \( dv = dy \) to get \( v = y \).
4Step 4: Apply Integration by Parts Formula
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int \tan^{-1}y \; dy = y \tan^{-1}y - \int y \cdot \frac{1}{1+y^2} dy \].
5Step 5: Simplify and Solve Remaining Integral
The remaining integral is \( \int \frac{y}{1+y^2} dy \). This can be solved by substitution. Let \( w = 1+y^2 \) then \( dw = 2y dy \) or \( \frac{dw}{2} = y dy \).
6Step 6: Solve Integral with Substitution
Substituting, the integral becomes \( \int \frac{1}{2} \frac{1}{w} dw \). This simplifies to \( \frac{1}{2} \ln |w| + C \). Substituting back, we have \( \frac{1}{2} \ln |1+y^2| + C \).
7Step 7: Combine Results
Substitute the result back into the integration by parts expression: \[ y \tan^{-1}y - \frac{1}{2} \ln |1+y^2| + C \].
Key Concepts
Inverse Trigonometric FunctionsDefinite Integral EvaluationSubstitution Method
Inverse Trigonometric Functions
Inverse trigonometric functions are functions that reverse the effects of the standard trigonometric functions such as sine, cosine, and tangent. These functions are denoted by the “arc” prefix or by the superscript (-1).
The most common inverse trigonometric functions include:
In the exercise, the function \(\tan^{-1}y\) is the inverse tangent of \(y\). This function is essential when working with integrals involving inverse trigonometric identities. It's especially useful in integration techniques like integration by parts.
Understanding the properties of inverse trigonometric functions is key:
The most common inverse trigonometric functions include:
- Arcsine (\(\sin^{-1}(x)\)): This function gives the angle whose sine is \(x\).
- Arccosine (\(\cos^{-1}(x)\)): Gives the angle whose cosine is \(x\).
- Arctangent (\(\tan^{-1}(x)\) or \(arctan(x)\)): Gives the angle whose tangent is \(x\).
In the exercise, the function \(\tan^{-1}y\) is the inverse tangent of \(y\). This function is essential when working with integrals involving inverse trigonometric identities. It's especially useful in integration techniques like integration by parts.
Understanding the properties of inverse trigonometric functions is key:
- They are often used to determine angles in calculus problems.
- They have derivatives that are particularly useful when evaluating integrals.
Definite Integral Evaluation
Definite integral evaluation is the process of finding the area under a curve between two specified points.
In cases like our example, although we are actually dealing with an indefinite integral, obtaining the antiderivative is the primary goal.
When evaluating integrals involving inverse trigonometric functions, it often requires specific techniques such as integration by parts. This is because inverse trigonometric functions do not always have straightforward antiderivatives, making them challenging.
To evaluate an indefinite integral, one finds the antiderivative (a general solution without specific bounds). This involves:
When evaluating integrals involving inverse trigonometric functions, it often requires specific techniques such as integration by parts. This is because inverse trigonometric functions do not always have straightforward antiderivatives, making them challenging.
To evaluate an indefinite integral, one finds the antiderivative (a general solution without specific bounds). This involves:
- Identifying the function to be integrated.
- Applying suitable integration techniques (like substitution or integration by parts).
- Combining any constants or additional functions that arise during integration.
Substitution Method
The substitution method, also known as the "u-substitution," simplifies complex integrals by changing the variable of integration. It's similar to reverse chain rule for differentiation.
In our solved example, substitution helps solve the integral \(\int \frac{y}{1+y^2} dy\). Here's how it works:
Substituting back to the original variable gives \(\frac{1}{2} \ln |1+y^2|\).
This method is fundamental for integrals that resist simple antiderivation, allowing one to turn complex expressions into simpler, solvable forms.
In our solved example, substitution helps solve the integral \(\int \frac{y}{1+y^2} dy\). Here's how it works:
- Choose a substitution that simplifies the integral. Let \(w = 1 + y^2\). This choice simplifies the denominator.
- Derive \(dw = 2y \, dy\) to replace \(y \, dy\) in the integral.
- Substitute and solve: The integral becomes \(\int \frac{1}{2w} dw\).
- Integration of \(\frac{1}{2w}\) yields an antiderivative \(\frac{1}{2} \ln |w|\).
Substituting back to the original variable gives \(\frac{1}{2} \ln |1+y^2|\).
This method is fundamental for integrals that resist simple antiderivation, allowing one to turn complex expressions into simpler, solvable forms.
Other exercises in this chapter
Problem 7
Use the table of integrals at the back of the book to evaluate the integrals. \(\int \frac{\sqrt{9-4 x}}{x^{2}} d x\)
View solution Problem 7
Evaluate the integrals in Exercises \(1-14\) $$ \int_{0}^{\pi} 8 \sin ^{4} x d x $$
View solution Problem 7
Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{d x}{\sqrt{x}(\sqrt{x}+1)} $$
View solution Problem 8
Evaluate the integrals in Exercises \(1-28\). $$ \int \sqrt{1-9 t^{2}} d t $$
View solution