Problem 34
Question
Evaluating the integral \(\int \arccos x d x\) requires two steps. (a) Write $$ \arccos x=1 \cdot \arccos x $$ and integrate by parts once to show that $$ \int \arccos x d x=x \arccos x+\int \frac{x}{\sqrt{1-x^{2}}} d x $$ (b) Use substitution to compute $$ \int \frac{x}{\sqrt{1-x^{2}}} d x $$ and combine your result in (a) with (7.12) to complete the computation of \(\int \arccos x d x\).
Step-by-Step Solution
Verified Answer
\(\int \arccos x \, dx = x \arccos x - \sqrt{1-x^2} + C\).
1Step 1: Setup the Integration by Parts
To evaluate the integral \(\int \arccos x \, dx\), write \(\arccos x = 1 \cdot \arccos x\). Here, choose \(u = \arccos x\) (thus, \(du = -\frac{1}{\sqrt{1-x^{2}}} \, dx\)) and \(dv = dx\) (thus, \(v = x\)). The integration by parts formula is \(\int u \, dv = uv - \int v \, du\).
2Step 2: Apply the Integration by Parts Formula
Substitute \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula: \(\int \arccos x \, dx = x \arccos x - \int x \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) \, dx\). Simplify to get \(x \arccos x + \int \frac{x}{\sqrt{1-x^{2}}} \, dx\). This shows \(\int \arccos x \, dx = x \arccos x + \int \frac{x}{\sqrt{1 - x^2}} \, dx\).
3Step 3: Setup the Substitution for Remaining Integral
To evaluate \(\int \frac{x}{\sqrt{1-x^{2}}} \, dx\), use a substitution. Let \(z = 1 - x^2\), hence \(dz = -2x \, dx\), or equivalently \(dx = -\frac{dz}{2x}\). Substituting into the integral gives \(\int \frac{x}{\sqrt{z}} \left(-\frac{dz}{2x}\right)\). After simplifying the expression, notice the cancellation of \(x\), which simplifies further to \(-\frac{1}{2} \int z^{-1/2} \, dz\).
4Step 4: Integrate and Combine Results
The integral \(-\frac{1}{2} \int z^{-1/2} \, dz\) is \(-\sqrt{z} + C\). Substitute back \(z = 1 - x^2\) to get \(-\sqrt{1-x^2} + C\). Combine this result with the expression from Step 2: \(\int \arccos x \, dx = x \arccos x + \left(-\sqrt{1-x^2}\right) + C\). Simplify the expression to get the final answer.
Key Concepts
Integration by PartsDefinite and Indefinite IntegralsTrigonometric FunctionsSubstitution Method
Integration by Parts
One vital concept in calculus is Integration by Parts, a technique that simplifies complex integrals into more manageable forms. It is often compared to the product rule of differentiation but in reverse. The Integration by Parts formula states that \( \int u \, dv = uv - \int v \, du \). Here, \(u\) and \(dv\) are chosen based on the integral you want to solve, and finding their derivatives and integrals respectively.
For the integral \( \int \arccos x \, dx \), we use a clever trick by setting \(u = \arccos x\) and \(dv = dx\). This means \(du = -\frac{1}{\sqrt{1-x^2}} \, dx\) and \(v = x\). Applying the formula simplifies the integral into two separate terms. Thus, the integral becomes \( x \arccos x + \int \frac{x}{\sqrt{1-x^2}} \, dx \). The choice of \(u\) and \(dv\) is essential as it determines how tractable the remaining integral will be.
For the integral \( \int \arccos x \, dx \), we use a clever trick by setting \(u = \arccos x\) and \(dv = dx\). This means \(du = -\frac{1}{\sqrt{1-x^2}} \, dx\) and \(v = x\). Applying the formula simplifies the integral into two separate terms. Thus, the integral becomes \( x \arccos x + \int \frac{x}{\sqrt{1-x^2}} \, dx \). The choice of \(u\) and \(dv\) is essential as it determines how tractable the remaining integral will be.
Definite and Indefinite Integrals
When dealing with integrals, distinguishing between definite and indefinite integrals is crucial. A definite integral calculates the area under a curve between two limits, while an indefinite integral represents a family of functions and includes a constant of integration, \(C\).
In the exercise at hand, we are focusing on the indefinite integral of \( \arccos x \). Evaluating indefinite integrals involves the Integration by Parts and substitution methods to simplify and compute their solutions. It's important to remember to add the constant \(C\) when you solve any indefinite integral, as this represents the set of all possible antiderivatives. After finalizing the integral terms, ensure everything is combined correctly to form the complete expression.
In the exercise at hand, we are focusing on the indefinite integral of \( \arccos x \). Evaluating indefinite integrals involves the Integration by Parts and substitution methods to simplify and compute their solutions. It's important to remember to add the constant \(C\) when you solve any indefinite integral, as this represents the set of all possible antiderivatives. After finalizing the integral terms, ensure everything is combined correctly to form the complete expression.
Trigonometric Functions
Trigonometric functions, like \(\arccos x\), play a significant role in calculus and algebra. These functions describe the relationships between the angles and sides of triangles. In this exercise, recognizing \(\arccos x\) as an inverse trigonometric function is key.
When dealing with these functions in integrals, it's important to understand their derivatives. The derivative of \(\arccos x\) is \(-\frac{1}{\sqrt{1-x^2}}\), which is used in the Integration by Parts process. Understanding the derivative helps simplify the integral and recognize patterns related to trigonometric identities, which can lead to easier computations and solutions.
When dealing with these functions in integrals, it's important to understand their derivatives. The derivative of \(\arccos x\) is \(-\frac{1}{\sqrt{1-x^2}}\), which is used in the Integration by Parts process. Understanding the derivative helps simplify the integral and recognize patterns related to trigonometric identities, which can lead to easier computations and solutions.
Substitution Method
The Substitution Method, often called \(u\)-substitution, transforms complex integrals into simpler ones by changing variables. It's similar to reversing the chain rule for derivatives, where you replace part of the integral with a new variable.
In this exercise, the substitution \(z = 1-x^2\) simplifies the integral \(\int \frac{x}{\sqrt{1-x^2}} \, dx\). From this substitution, \(dz = -2x \, dx\) leads to \(dx = -\frac{dz}{2x}\). This simplifies further when substituted back into the integral, converting it to \(-\frac{1}{2} \int z^{-1/2} \, dz\). Solving this using basic integration rules results in \(-\sqrt{z}\), which translates back to \(-\sqrt{1-x^2}\) when the substitution is reversed, completing the integral process effectively.
In this exercise, the substitution \(z = 1-x^2\) simplifies the integral \(\int \frac{x}{\sqrt{1-x^2}} \, dx\). From this substitution, \(dz = -2x \, dx\) leads to \(dx = -\frac{dz}{2x}\). This simplifies further when substituted back into the integral, converting it to \(-\frac{1}{2} \int z^{-1/2} \, dz\). Solving this using basic integration rules results in \(-\sqrt{z}\), which translates back to \(-\sqrt{1-x^2}\) when the substitution is reversed, completing the integral process effectively.
Other exercises in this chapter
Problem 33
In Problems 17-36, use substitution to evaluate each indefinite integral. $$ \int \frac{(\ln x)^{2}}{x} d x $$
View solution Problem 34
Determine the constant \(c\) so that $$ \int_{-\infty}^{\infty} \frac{c}{1+x^{2}} d x=1 $$
View solution Problem 34
Use partial-fraction decomposition to evaluate the integrals. $$ \int \frac{1}{(x-1)(x+2)} d x $$
View solution Problem 34
In Problems 17-36, use substitution to evaluate each indefinite integral. $$ \int \frac{d x}{(x+3) \ln (x+3)} $$
View solution