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Chapter 8: Problem 30

Use a substitution to change the integral into one you can find in the table. Then evaluate the integral. \(\int \frac{\cos ^{-1} \sqrt{x}}{\sqrt{x}} d x\)

Short Answer

Expert verified
The integral evaluates to \( 2\cos^{-1}(\sqrt{x})\sqrt{x} - 2\sqrt{1-x} + C \).

Step by step solution

01

Identify a suitable substitution

Notice that the integrand involves both an inverse cosine function and a square root function. Let's try the substitution \( u = \cos^{-1}(\sqrt{x}) \). Then, the differential \( du \) can be established from this substitution.
02

Express \( x \) in terms of \( u \) and find \( dx \)

From the equation \( u = \cos^{-1}(\sqrt{x}) \), we get \( \sqrt{x} = \cos(u) \). Squaring both sides gives \( x = \cos^2(u) \). Differentiating both sides with respect to \( u \), we find \( dx = -2\cos(u)\sin(u) \, du \).
03

Rewrite the integral in terms of \( u \)

Substitute back into the integral: \( \int \frac{\cos^{-1}(\sqrt{x})}{\sqrt{x}} \, dx = \int \frac{u}{\cos(u)} (-2\cos(u)\sin(u)) \, du \). Simplifying, this becomes \(-2 \int u \sin(u) \, du \).
04

Integrate using a known formula

The integral \( \int u \sin(u) \, du \) can be evaluated using integration by parts. Let \( v = u \) and \( dw = \sin(u) \, du \). This means \( dv = du \) and \( w = -\cos(u) \). Thus, the integral becomes \( -2\left[-u \cos(u) + \int \cos(u) \, du\right] \).
05

Solve the remaining simple integral

The integral \( \int \cos(u) \, du \) is simply \( \sin(u) \). So the solution to the integral in Step 4 is \( -2[-u \cos(u) + \sin(u)] = 2u\cos(u) - 2\sin(u) \).
06

Substitute back in terms of \( x \)

Recall that \( u = \cos^{-1}(\sqrt{x}) \), \( \cos(u) = \sqrt{x} \), and \( \sin(u) = \sqrt{1-x} \). Substitute these back into the integral expression to get: \( 2\cos^{-1}(\sqrt{x})\sqrt{x} - 2\sqrt{1-x} + C \), where \( C \) is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method, also known as u-substitution, is a technique used in integration to simplify an integral by changing its variable. It often turns a complicated integral into a form that is easier to evaluate. In this exercise, we started with an integral that involved an inverse trigonometric function and a square root. Using the substitution \( u = \cos^{-1}(\sqrt{x}) \) helped simplify this complex expression.Steps of Substitution Method:- Identify a part of the integrand that can be replaced with a simple variable \( u \).- Express other variables in terms of \( u \) and decide on the new differential \( du \).- Rewrite the original integral using \( u \) and \( du \).Substituting this way reduced the problem to a simpler integral in terms of \( u \), which was then easier to solve.
Inverse Trigonometric Functions
Inverse trigonometric functions are functions that reverse the action of the basic trigonometric functions like sine, cosine, and tangent. In our integral, the inverse cosine function, \( \cos^{-1} \), was initially used. This function gives an angle whose cosine is a specified value.Some common properties of inverse trigonometric functions:- They are the inverse operations of trigonometric functions.- They are often denoted with prefixes like \( \sin^{-1}, \tan^{-1}, \cos^{-1} \).In the exercise, knowing the property \( u = \cos^{-1}(\sqrt{x}) \) was essential, allowing substitution and simplification.
Integration by Parts
Integration by parts is a method applied to integrate products of functions. The formula is derived from the product rule of differentiation and is given by:\[ \int u \, dv = uv - \int v \, du \]For the given integral, \( -2 \int u \sin(u) \, du \), we used integration by parts by selecting \( u \) and \( dv \) appropriately:- Let \( u = u \) (the same variable) and \( dv = \sin(u) \, du \).- Then, \( du = du \), and \( v = -\cos(u) \).Applying integration by parts, the integral became easier to evaluate and allowed for solving the remaining parts.
Definite and Indefinite Integrals
Integrals in calculus can be of two types: definite and indefinite. An indefinite integral represents a family of functions and includes a constant of integration \( C \), while a definite integral computes the area under a curve and produces a numerical result.Key Differences:- **Indefinite Integrals:** Give an expression with \( + C \).- **Definite Integrals:** Provide a bounded area with specific limits.In this exercise, we calculated an indefinite integral since there were no upper or lower limits specified. After integrating, it was important to include the constant of integration \( C \) in the final result.

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Most popular questions from this chapter

The infinite paint can or Gabriel's horn As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x, $$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\) -axis, diverges also. By comparing the two integrals, we see that, for every finite value \(b>1\) , $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x.$$ However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. \begin{equation} \begin{array}{l}{\text { a. Calculate it. }} \\ {\text { b. This solid of revolution is sometimes described as a can that }} \\ \quad {\text { does not hold enough paint to cover its own interior. Think }} \\ \quad {\text { about that for a moment. It is common sense that a finite }} \\ \quad {\text { amount of paint cannot cover an infinite surface. But if we fill }} \\ \quad {\text { the horn with paint (a finite amount), then we will have covered }} \\\ \quad {\text { an infinite surface. Explain the apparent contradiction. }} \end{array} \end{equation}

Find the value of \(c\) so that \(f(x)=c \sqrt{x}(1-x)\) is a probability density function for the random variable \(X\) over \([0,1],\) and find the probability \(P(0.25 \leq X \leq 0.5)\)

Customer service time The The mean waiting time to get served after walking into a bakery is 30 seconds. Assume that an exponential density function describes the waiting times. $$ \begin{array}{l}{\text { a. What is the probability a customer waits } 15 \text { seconds or less? }} \\ {\text { b. What is the probability a customer waits longer than one }} \\ {\text { minute? }} \\ {\text { c. What is the probability a customer waits exactly } 5 \text { minutes? }} \\ {\text { d. If } 200 \text { customers come to the bakery in a day, how many are }} \\\ {\text { likely to be served within three minutes? }}\end{array} $$

Prove that the sum \(T\) in the Trapezoidal Rule for \(\int_{a}^{b} f(x) d x\) is a Riemann sum for \(f\) continuous on \([a, b] .\) Hint: Use the Intermediate Value Theorem to show the existence of \(c_{k}\) in the subinterval \(\left[x_{k-1}, x_{k}\right]\) satisfying \(f\left(c_{k}\right)=\left(f\left(x_{k-1}\right)+f\left(x_{k}\right)\right) / 2 . )\)

Use numerical integration to estimate the value of $$ \sin ^{-1} 0.6=\int_{0}^{0.6} \frac{d x}{\sqrt{1-x^{2}}} $$ For reference, \(\sin ^{-1} 0.6=0.64350\) to five decimal places.
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