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Chapter 4: Problem 7

Evaluate the indicated integral. $$\begin{aligned} &4\\\ &\int \frac{\sin x}{\sqrt{\cos x}} d x \end{aligned}$$

Short Answer

Expert verified
The integral of the given function is \(-8\sqrt{\cos x}+C\)

Step by step solution

01

Substitute

Firstly, let \(u = \cos x\). Then, \(du = -\sin x~dx\). The negative sign can be adjusted by multiplying by -1.
02

Substitute variable u into integral and adjust the negative sign

After the substitution, the integral becomes \(-4\int \frac{1}{\sqrt{u}}~du\). Now, the integral can be simplified using the power rule.
03

Simplify using the power rule

Using the power rule, the integral now becomes \(-4(2u^\frac{1}{2})+C\).
04

Substitute u back

As the last step, substitute u back into the result. So, \(-4(2u^\frac{1}{2})+C\) becomes \(-4(2\sqrt{\cos x})+C\).\nit's simplified as \(-8\sqrt{\cos x}+C\).

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Key Concepts

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

Integration by Substitution
Integration by substitution is a crucial technique in integral calculus. Think of it as a method for simplifying complex integrals into more manageable forms. It's akin to changing the variables to a new set that simplifies the problem.

Here's how it works:
  • Firstly, identify a portion of the integral that, when substituted with a new variable, simplifies the integral. In our example, we chose the substitution \( u = \cos x \).
  • Next, differentiate your chosen substitution to find \( du \), which is the derivative of \( u \) with respect to \( x \). In the example, \( du = -\sin x \, dx \).
  • The negative sign here is adjusted by multiplying by -1. This is important since it helps when arranging the integrand in terms of \( du \).
  • Replace all instances of \( x \) in the integral with your substitution \( u \). This transforms the original integral into one in terms of \( u \), often making it simpler to integrate.
This technique can be particularly useful when you're dealing with trigonometric functions, as they often have derivatives that fit well into the substitution framework.
Trigonometric Integrals
Trigonometric integrals involve integrating functions that contain trigonometric expressions like \(\sin x\), \(\cos x\), \(\tan x\), and so on. They frequently require substitution and identities to simplify.Let's explore how they work:
  • Many trigonometric integrals are approached by using identities to rewrite the expression in a different form. For instance, using identities like \(\sin^2 x + \cos^2 x = 1\) can be immensely helpful.
  • Prior experiences hinting at patterns or familiar results often aid in transforming or simplifying trigonometric expressions.
  • Substitution can further simplify these integrals, as seen previously. Here, the integral \(\int \frac{\sin x}{\sqrt{\cos x}} \, dx\) simplifies greatly with the choice \( u = \cos x \), making it ready for application of power rules.
Mastering trigonometric integrals entails comfort with substitutions and trigonometric identities, creating a smoother path to finding integrals of trigonometric functions.
Power Rule
The power rule in calculus is a fundamental principle for integration. It allows you to integrate expressions of the form \( x^n \), where \( n \) is any real number, by systematically following a formula.Here's how it applies:
  • The generic formula for integration using the power rule is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.
  • When applying this rule to expressions with fractional exponents, such as those that arise in substitution, care must be taken to adjust the expression accordingly.
  • In our specific problem, after substituting besides \( u = \cos x \), you simplify the expression to a form suitable for the power rule, i.e., \(-4\int u^{-1/2} \, du \). Applying the power rule then yields \(-4(2u^{1/2}) + C \).
The power rule is a powerful tool, especially when combined with substitution methods, as it directly leads to the integral's evaluation. Understanding it is foundational for tackling more complex integrals effectively.

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Most popular questions from this chapter

Show that \(\int \frac{-1}{\sqrt{1-x^{2}}} d x=\cos ^{-1} x+c\) and \(\int \frac{-1}{\sqrt{1-x^{2}}} d x=-\sin ^{-1} x+c\) Explain why this does not imply that \(\cos ^{-1} x=-\sin ^{-1} x .\) Find an equation relating \(\cos ^{-1} x\) and \(\sin ^{-1} x\)

Suppose you have a 1 -in- 10 chance of winning a prize with some purchase (like a lottery). If you make 10 purchases (i.e., you get 10 tries), the probability of winning at least one prize is \(1-(9 / 10)^{10} .\) If the prize had probability 1 -in-20 and you tried 20 times, would the probability of winning at least once be higher or lower? Compare \(1-(9 / 10)^{10}\) and \(1-(19 / 20)^{20}\) to find out. To see what happens for larger and larger odds, compute \(\lim _{n \rightarrow \infty}\left[1-((n-1) / n)^{n}\right].\)

Involve the just-in-time inventory discussed in the chapter introduction. The EOQ model of exercise 62 can be modified to take into account noninstantaneous receipt. In this case, instead of a full delivery arriving at one instant, the delivery arrives at a rate of \(p\) items per day. Assume that a delivery of size \(Q\) starts at time \(0,\) with shipments out continuing at the rate of \(r\) items per day (assume that \(p>r\) ). Show that when the delivery is completed, the inventory equals \(Q(1-r / p) .\) From there, inventory drops at a steady rate of \(r\) items per day until no items are left. Show that the average inventory equals \(\frac{1}{2} Q(1-r / p)\) and find the order size \(Q\) that minimizes the total cost.

The location \((\bar{x}, \bar{y})\) of the center of gravity (balance point) of a flat plate bounded by \(y=f(x)>0, a \leq x \leq b\) and the \(x\) -axis is given by \(\bar{x}=\frac{\int_{a}^{b} x f(x) d x}{\int_{a}^{b} f(x) d x}\) and \(\bar{y}=\frac{\int_{a}^{b}[f(x)]^{2} d x}{2 \int_{a}^{b} f(x) d x} .\) For the semicircle \(y=f(x)=\sqrt{4-x^{2}},\) use symmetry to argue that \(\bar{x}=0\) and \(\bar{y}=\frac{1}{2 \pi} \int_{0}^{2}\left(4-x^{2}\right) d x .\) Compute \(\bar{y}\)

In most of the calculations that you have done, it is true that the Trapezoidal Rule and Midpoint Rule are on opposite sides of the exact integral (i.e., one is too large, the other too small). Also, you may have noticed that the Trapezoidal Rule tends to be about twice as far from the exact value as the Midpoint Rule.Given this, explain why the linear combination \(\frac{1}{3} T_{n}+\frac{2}{3} M_{n}\) should give a good estimate of the integral. (Here, \(T_{n}\) represents the Trapezoidal Rule approximation using \(n\) partitions and \(M_{n}\) the corresponding Midpoint Rule approximation.)
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