/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Find the general antiderivative ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the general antiderivative of the given function. $$ f(x)=\cos ^{2} x+1 $$

Short Answer

Expert verified
The general antiderivative is \( F(x) = \frac{3}{2}x + \frac{1}{4}\sin(2x) + C \).

Step by step solution

01

Identify the structure of the function

The given function is composed of two parts: a trigonometric function squared, \( \cos^2 x \), and a constant, 1. Our task is to find the antiderivative of the entire function.
02

Recall the identity for trigonometric power

Use the trigonometric identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \). This identity helps simplify \( \cos^2 x \) into a more integrable form.
03

Rewrite the function using the identity

Substitute the identity into the function: \( f(x) = \frac{1 + \cos(2x)}{2} + 1 \). This simplifies to \( f(x) = \frac{1}{2} + \frac{\cos(2x)}{2} + 1 \).
04

Combine like terms

Simplify the function further: \( f(x) = \frac{3}{2} + \frac{\cos(2x)}{2} \). Separate this expression for easy integration: \( f(x) = \frac{3}{2} + \frac{1}{2}\cos(2x) \).
05

Integrate each term separately

Find the antiderivative of each term:- The antiderivative of \( \frac{3}{2} \) is \( \frac{3}{2}x \).- The antiderivative of \( \frac{1}{2}\cos(2x) \) is \( \frac{1}{2} \times \frac{1}{2} \sin(2x) = \frac{1}{4}\sin(2x) \), using the substitution rule and dividing by the coefficient of \( x \) in the cosine argument.
06

Combine the antiderivatives

The general antiderivative of \( f(x) = \cos^2 x + 1 \) is the combination of the antiderivatives from Step 5: \( F(x) = \frac{3}{2}x + \frac{1}{4}\sin(2x) + C \), 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.

Trigonometric Functions
Trigonometric functions are fundamental in calculus, particularly when dealing with periodic functions such as sine and cosine. In our exercise, the function includes \( \cos^2 x \), a common trigonometric function that often requires simplification to integrate effectively.
To work with trigonometric integrals, we frequently use identities that simplify the functions involved. For instance, the power-reduction identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \) is powerful.
It transforms a squared trigonometric function into a form that is easier to integrate. This identity allows us to rewrite \( \cos^2 x \) into expressions of half-angles which results in simpler integrands.
Understanding and applying trigonometric identities helps in efficiently breaking down complex functions into simpler parts, setting the stage for successful integration.
Integration Techniques
Mastering integration techniques is essential for solving a wide array of calculus problems, including finding antiderivatives of complex functions. For the expression \( f(x) = \frac{3}{2} + \frac{1}{2} \cos(2x) \), we encounter a straightforward process due to its simpler structure.
When integrating, two main techniques are:
  • Direct Integration: If the function is a sum of simple terms like a constant or basic trigonometric expression, it can be integrated term by term.
  • Using Identities: Trigonometric identities, like in this problem, assist in breaking down expressions into more integrable forms.
Each term of the given function can be dealt with separately.
The antiderivative of a constant \( \frac{3}{2} \) is simply \( \frac{3}{2}x \), while for \( \frac{1}{2}\cos(2x) \), we rely on the substitution method to handle the cosine with a linear argument.
Substitution Method
The substitution method is a powerful technique for integrating functions where the integrand involves a composite function. It simplifies this process by converting the integral of a complex function into an easier, standard form.
In our exercise, this method plays a crucial role when integrating the term \( \frac{1}{2} \cos(2x) \).
Substitution is made by recognizing that the inner function, \( 2x \), requires us to compensate by dividing by the derivative of \( 2x \), which is 2. Hence the integral of \( \cos(2x) \) becomes \( \frac{1}{2} \sin(2x) \), and with the additional coefficient, the term is fully integrated to \( \frac{1}{4}\sin(2x) \).
The key steps in substitution include:
  • Identifying the inner function and its derivative.
  • Adjusting the integral accordingly by compensating for the derivative.
This method not only aids in quicker calculations but also in understanding the underlying structure of composite functions in calculus.

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Most popular questions from this chapter

In Problem 9 we neglected to consider the time delay between a pill being taken and the drug entering the patient's blood. In Chapter 8 we will introduce compartment models as models for drug absorption. We will show that a good model for a drug being absorbed from the gut is that the rate of drug absorption, \(A(t)\), varies with time according to: $$ A(t)=C e^{-k t}, t \geq 0 $$ where \(C>0\) and \(k>0\) are coefficients that will depend on the type of drug, as well as varying between patients. (a) Assume that the drug has first order elimination kinetics, with elimination rate \(k_{1} .\) Show that the amount of drug in the patient's blood will obey a differential equation: $$ \frac{d M}{d t}=C e^{-k t}-k_{1} M $$ (b) Verify that a solution of this differential equation is: $$ M(t)=\frac{C e^{-k t}}{k_{1}-k}+a e^{-k_{1} t} $$ where \(a\) is any coefficient, and we assume \(k_{1} \neq k\). (c) To determine the coefficient \(a\), we need to apply an initial condition. Assume that there was no drug present in the patient's blood when the pill first entered the gut (that is, \(M(0)=0\) ). Find the value of \(a\). (d) Let's assume some specific parameter values. Let \(C=2\), \(k=3\), and \(k_{1}=1 .\) Show that \(M(t)\) is initially increasing, and then starts to decrease. Find the maximum level of drug in the patient's blood. (e) Show that \(M(t) \rightarrow 0\) as \(t \rightarrow \infty\). (f) Using the information from (d) and (e), make a sketch of \(M(t)\) as a function of \(t\).

Use the Newton-Raphson method to find a numerical approximation for all of the solutions of: $$ x^{4}+x^{3}+1=x^{2}+2 x $$ correct to six decimal places.

Elimination of ethanol from the blood is known to have zeroth order kinetics. Provided no more ethanol enters the blood, the concentration of ethanol in a person's blood will therefore obey the following differential equation: $$ \frac{d M}{d t}=-k_{0} $$ where for a typical adult \(k_{0}=0.186 \mathrm{~g} /\) liter \(/ \mathrm{hr}\) (al-Lanqawi et al. 1992). (a) Explain why \(M(t)\) can only obey the above differential equation if \(M>0\) (once \(M\) drops to 0 , it is usual to assume that \(\left.\frac{d M}{d t}=0\right)\) (b) If a person's blood alcohol concentration is \(1.6 \mathrm{~g}\) /liter at midnight, what will their blood alcohol concentration be at \(2 \mathrm{am}\) ? You may assume that she drinks no more alcohol after midnight. (c) At what time will their blood alcohol concentration drop to \(0 \mathrm{~g} /\) liter?

A generalization of the Beverton-Holt model for population growth was created by Hassell (1975). Under Hassell's model the population \(N_{t}\) at discrete times \(t=0,1,2, \ldots\) is modeled by a recurrence equation: $$N_{t+1}=\frac{R_{0} N_{t}}{\left(1+a N_{t}\right)^{c}}$$ where \(R_{0}, a\), and \(c\) are all positive constants. (a) Explain why you would expect that \(R_{0}>1\). (b) Assume that \(c=2, R_{0}=9\), and \(a=\frac{1}{10} .\) Find all possible equilibria of the system. (c) Use the stability criterion for equilibria to determine which, if any, of the equilibria of the recursion relation are stable.

Use the Newton-Raphson method to find a numerical approximation to the solution of $$ x^{2}-7=0 $$ that is correct to six decimal places.

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