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Use a table of integrals to determine the following indefinite integrals. These integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table. $$\int \frac{\cos ^{-1} \sqrt{x}}{\sqrt{x}} d x$$

Short Answer

Expert verified
Question: Find the indefinite integral of the given function: $$\int\frac{\cos^{-1}\sqrt{x}}{\sqrt{x}}dx$$ Answer: $$\int\frac{\cos^{-1}\sqrt{x}}{\sqrt{x}}dx=2(\cos^{-1}\sqrt{x})\sqrt{x}+\sqrt{1-x}+C$$

Step by step solution

01

Identify the substitution needed

In this problem, it is helpful to consider the substitution: $$ u = \cos^{-1}\sqrt{x} $$ which gives: $$ x = \cos^2(u) $$
02

Find the derivative of "u"

Differentiate "\(u\)" with respect to "x": $$ \frac{d u}{d x} = \frac{-1}{2\sqrt{x}\sqrt{1-x}} $$
03

Find the derivative of "x" with respect to "u"

Now find the derivative of "x" with respect to "u": $$ \frac{d x}{d u} = -2\cos(u)\sin(u)$$
04

Rewrite the integral in terms of "u"

Now we can rewrite the integral as follows: $$ \int \frac{\cos ^{-1} \sqrt{x}}{\sqrt{x}} dx = \int \frac{u}{\sqrt{\cos^2(u)}}(-2\cos(u)\sin(u))du $$ Simplify the right-hand side expression: $$ \int -2u\sin(u)du $$
05

Use integration by parts

Integration by parts formula: $$ \int u dv = uv - \int v du $$ Choose "u" and "dv": $$ u = -2u, \hspace{1cm} dv = \sin(u)du $$ Find "du" and "v": $$ du = -2du, \hspace{1cm} v = -\cos(u) $$ Apply integration by parts: $$ \int -2u\sin(u)du = (-2u)(-\cos(u)) - \int (-\cos(u))(-2du) $$
06

Simplify and integrate

Simplify the equation and integrate: $$ 2u\cos(u) - 2\int \cos(u)du $$ Integrate the remaining part: $$ 2u\cos(u) + 2\sin(u) + C $$
07

Substitute back to original variable "x"

Finally, substitute the original variable, "x", back into the equation using the substitution we found in Step 1: $$ 2(\cos^{-1}\sqrt{x})\cos(\cos^{-1}\sqrt{x}) + 2\sin(\cos^{-1}\sqrt{x}) + C $$
08

Simplify the result

Simplify the expression: $$ 2(\cos^{-1}\sqrt{x})\sqrt{x}+\sqrt{1-x} + C $$ And now we have the indefinite integral of the given function: $$ \boxed{\int\frac{\cos^{-1}\sqrt{x}}{\sqrt{x}}dx=2(\cos^{-1}\sqrt{x})\sqrt{x}+\sqrt{1-x}+C} $$

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Key Concepts

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

Indefinite Integrals
Indefinite integrals refer to the collection of all possible antiderivatives of a given function. Unlike definite integrals, which produce a numerical value, indefinite integrals generate a function plus an arbitrary constant, represented by "C". This constant accounts for the infinite number of antiderivatives a function can have. For example, if you compute the indefinite integral of a function like \(f(x)\), you aim to find a function \(F(x)\) such that \(F'(x) = f(x)\). Basic knowledge of antiderivatives allows you to tackle more complex integration problems by breaking them down into simpler parts or applying specific techniques like substitution and integration by parts.Indefinite integrals play a crucial role in finding the area under curves and in solving differential equations, making them a fundamental aspect of calculus and higher mathematics.
Trigonometric Substitution
Trigonometric substitution is a technique used in calculus to simplify the integration of functions involving square roots. This method involves replacing variables with trigonometric functions, typically sine, cosine, or tangent, to make the integral more manageable. When you encounter an integral with a square root like \( \sqrt{x} \), you might substitute \( x = \cos^2(u) \) to simplify the expression.The essence of trigonometric substitution is in transforming the given variable into a function of an angle \(u\), where trigonometric identities can significantly simplify mathematical expressions. During the process, you calculate derivatives or use identities like \( \sin^2(u) + \cos^2(u) = 1 \) to aid in conversion and simplification.Utilizing trigonometric substitution not only helps in solving complex integrals but also provides a deeper understanding of how trigonometric functions can be applied beyond trigonometry itself.
Integration by Parts
Integration by parts is a powerful technique derived from the product rule of differentiation, used to integrate the product of two functions. This method is particularly useful when dealing with products of algebraic and trigonometric, exponential, or logarithmic functions. The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]To apply this method, you first need to identify parts of the integrand to represent \( u \) and \( dv \). Then, differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\). By substituting these into the integration by parts formula, you can often transform a complex integral into a more straightforward one.Choosing \( u \) and \( dv \) properly is key to effectively using integration by parts. The acronym "LIATE" (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) can guide you in deciding the order of preference for choosing \(u\). As you practice this technique, it becomes an invaluable tool in your integration arsenal.
Calculus
Calculus is a branch of mathematics focused on change and motion, based on the concepts of derivatives and integrals. This field is divided into two main branches: differential calculus and integral calculus. Differential calculus revolves around the concept of derivatives, which describe how quantities change. Integral calculus, on the other hand, focuses on integrals, dealing with accumulation of quantities and areas under curves. The fundamental theorem of calculus links these two branches, showing how differentiation and integration are inverse processes. This theorem is pivotal as it simplifies the evaluation of definite integrals by using antiderivatives. Calculus plays a critical role in numerous fields such as physics, engineering, economics, statistics, and many more. It allows us to model real-world situations and solve problems related to rates of change, optimization, and motion, thereby offering extensive practical applications across various disciplines.

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Most popular questions from this chapter

Prove that the Trapezoid Rule is exact (no error) when approximating the definite integral of a linear function.

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}.\) a. Find the work required to launch an object in terms of \(m.\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x}$$

Refer to Theorem 2 and let \(f(x)=\sin e^{x}\) a. Find a Trapezoid Rule approximation to \(\int_{0}^{1} \sin \left(e^{x}\right) d x\) using \(n=40\) subintervals. b. Calculate \(f^{\prime \prime}(x)\) c. Explain why \(\left|f^{\prime \prime}(x)\right|<6\) on \([0,1],\) given that \(e<3\). (Hint: Graph \(\left.f^{\prime \prime} .\right)\) d. Find an upper bound on the absolute error in the estimate found in part (a) using Theorem 2.

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$ F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t $$ where we assume that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=e^{a t} \longrightarrow F(s)=\frac{1}{s-a}$$

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