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Chapter 7: Problem 37

Find the integrals Check your answers by differentiation. $$\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$$

Short Answer

Expert verified
The integral is \( 2 \sin(\sqrt{x}) + C \). Differentiating it gives \( \frac{\cos(\sqrt{x})}{\sqrt{x}} \), confirming correctness.

Step by step solution

01

Identify the Integration Technique

To solve the integral \( \int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx \), we need to identify an appropriate integration technique. The integral suggests using a substitution method, specifically aimed at simplifying the expression under the integral.
02

Choose the Substitution

Set \( u = \sqrt{x} \), which implies that \( x = u^2 \). Therefore, \( dx = 2u \, du \) when differentiating both sides. We can now rewrite the integral in terms of \( u \):
03

Rewrite the Integral

Substituting \( u = \sqrt{x} \) and \( dx = 2u \, du \), the integral becomes: \[\int \frac{\cos u}{u} \cdot 2u \, du = 2 \int \cos u \, du\]The \( u \) terms cancel, leaving a simpler integral to solve.
04

Integrate with Respect to u

Integrate \( 2 \int \cos u \, du \): \[2 \int \cos u \, du = 2 \sin u + C\]where \( C \) is the constant of integration. This is the integrated result in terms of \( u \).
05

Substitute Back

Replace \( u \) with \( \sqrt{x} \) to write the final answer in terms of \( x \):\[2 \sin(\sqrt{x}) + C\]This represents the integral in terms of the original variable.
06

Verify by Differentiation

Differentiate \( 2 \sin(\sqrt{x}) + C \) with respect to \( x \) to verify our result. Apply chain rule differentiation:\( \frac{d}{dx} [2 \sin(\sqrt{x}) + C] = 2 \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\cos(\sqrt{x})}{\sqrt{x}} \).The derivative matches the original integrand, confirming the result.

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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 is a popular technique in calculus, often used to simplify complicated integrals. Here's how it works:
  • You begin by choosing a new variable, typically denoted as \( u \), to replace part of the original integral. The goal is to turn the integral into a simpler one.
  • This involves substituting a function of \( x \), often something inside a more complex function or expression, with the variable \( u \). The variable \( x \) is expressed in terms of \( u \); for example, in our exercise \( u = \sqrt{x} \).
  • After substituting, you must convert \( dx \) into terms of \( du \) using differentiation. In the exercise, set \( dx = 2u \, du \) to align the new integral with the \( u \) variable.
This approach breaks down the integration problem, simplifying the algebra involved so the integral becomes easier to evaluate. Once integrated with respect to \( u \), you switch back to the original variable \( x \) by substituting back the initial function for \( u \). This ensures the solution is presented in terms of the variable used in the original problem.
Definite Integrals
Definite integrals are crucial in understanding the area under a curve or finding the accumulation of a quantity. Unlike indefinite integrals, which include an arbitrary constant \( C \), a definite integral calculates a specific numerical value by integrating between two limits. Here are some key points:
  • They are expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
  • To solve a definite integral, first find the indefinite integral, then evaluate it at the upper limit \( b \) and subtract the value at the lower limit \( a \).
  • The result corresponds to the signed area between the graph of the function \( f(x) \) and the \( x \)-axis over the interval \([a, b]\).
Although the exercise we covered is an indefinite integral (since it includes the constant \( C \)), understanding definite integrals is essential for complete comprehension of integration. They find extensive use in physics, engineering, and various applications where specific net values or totals are required.
Chain Rule
The chain rule is a fundamental calculus concept used to differentiate composite functions. It plays a vital role in checking our integration results through differentiation.
  • When you have a function nested inside another function, like \( \sin(\sqrt{x}) \), you need to apply the chain rule to differentiate it.
  • The chain rule states that the derivative \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). Here, \( f \) is the outer function and \( g \) is the inner function.
  • For the exercise, to verify the integral \( 2\sin(\sqrt{x}) \), differentiate using the chain rule: \( 2\cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \). This simplifies directly to our original integrand \( \frac{\cos(\sqrt{x})}{\sqrt{x}} \).
The chain rule allows for effective differentiation of complex composite functions, confirming the integration result is accurate. Applying it correctly is crucial to mastering calculus.

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Most popular questions from this chapter

(a) Between 2005 and \(2015,\) ACME Widgets sold widgets at a continuous rate of \(R=R_{0} e^{0.125 t}\) widgets per year, where \(t\) is time in years since January 1 2005\. Suppose they were selling widgets at a rate of 1000 per year on January \(1,2005 .\) How many widgets did they sell between 2005 and \(2015 ?\) How many did they sell if the rate on January 1,2005 was 1,000,000 widgets per year? (b) In the first case ( 1000 widgets per year on January 1,2005)\(,\) how long did it take for half the widgets in the ten-year period to be sold? In the second case \((1.000 .000 \text { widgets per year on January } 1,2005)\) when had half the widgets in the ten-year period been sold? (c) In \(2015,\) ACME advertised that half the widgets it had sold in the previous ten years were still in use. Based on your answer to part (b), how long must a widget last in order to justify this claim?

Find the exact area of the regions.Bounded by \(y=\left(3 x^{2}+x\right) /\left(\left(x^{2}+1\right)(x+1)\right)\) \(y=0, x=0, x=1\).

Give an example of:A linear polynomial \(P(x)\) and a quadratic polynomial \(Q(x)\) such that the rational function \(P(x) / Q(x)\) does not have a partial fraction decomposition of the form $$\frac{P(x)}{Q(x)}=\frac{A}{x-r}+\frac{B}{x-s}$$,for some constants \(A, B, r,\) and \(s\).

Decide whether the statements are true or false. Give an explanation for your answer.The integral \(\int \frac{1}{\sqrt{9-t^{2}}} d t\) can be made easier to evaluate by using the substitution \(t=3 \tan \theta\).

Find an expression for the integral which contains \(g\) but no integral sign. $$\int g^{\prime}(x) \sin g(x) d x$$
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