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Chapter 5: Problem 42

Evaluate the integrals using appropriate substitutions. $$ \int \frac{\cos 4 \theta}{(1+2 \sin 4 \theta)^{4}} d \theta $$

Short Answer

Expert verified
\(-\frac{1}{24(1 + 2\sin 4\theta)^3} + C\)

Step by step solution

01

Identify the substitution

First, let's identify a substitution that can simplify the integrand. A common approach when dealing with trigonometric functions is to use the substitution related to the argument of the sine or cosine function. Here, since we have both \( \cos 4\theta \) and \( \sin 4\theta \), we'll use the substitution \( u = 1 + 2 \sin 4\theta \).
02

Compute the derivative of the substitution

Next, find \( \frac{du}{d\theta} \) to express \( d\theta \) in terms of \( du \). From the substitution \( u = 1 + 2 \sin 4\theta \), differentiate both sides:\[\frac{du}{d\theta} = 2 \cdot 4 \cos 4\theta = 8 \cos 4\theta\]Thus, \( du = 8 \cos 4\theta \, d\theta \), or equivalently \( d\theta = \frac{du}{8 \cos 4\theta} \).
03

Rewrite the integrand using the substitution

Substitute \( u = 1 + 2 \sin 4\theta \) and \( d\theta = \frac{du}{8 \cos 4\theta} \) into the integral:\[\int \frac{\cos 4\theta}{(1 + 2 \sin 4\theta)^4} \, d\theta = \int \frac{\cos 4\theta}{u^4} \cdot \frac{du}{8 \cos 4\theta}\]The \( \cos 4\theta \) terms cancel out, simplifying to:\[\int \frac{1}{8u^4} \, du\]
04

Integrate with respect to \( u \)

Now, the integral is much simpler:\[\int \frac{1}{8u^4} \, du = \frac{1}{8} \int u^{-4} \, du\]This integral is a basic power integral and equals:\[\frac{1}{8} \cdot \left( \frac{u^{-3}}{-3} \right) = -\frac{1}{24} u^{-3} + C\]So, we have:\[-\frac{1}{24} \cdot \frac{1}{u^3} + C\]
05

Substitute back to \( \theta \)

Re-substitute \( u = 1 + 2 \sin 4\theta \) back into the solution:\[-\frac{1}{24} \cdot \frac{1}{(1 + 2 \sin 4\theta)^3} + C\]This is the final result of the integration in terms of \( \theta \).

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Key Concepts

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

Trigonometric substitution
Trigonometric substitution is a powerful technique in integral calculus that can simplify integrals involving trigonometric functions. The idea is to exploit the trigonometric identities and relationships to transform a potentially complex integral into a simpler form. In our exercise, we started with the integral \[ \int \frac{\cos 4 \theta}{(1+2 \sin 4 \theta)^{4}} d \theta \]by choosing a clever substitution. The substitution we used was \( u = 1 + 2 \sin 4\theta \).
  • First, identify expressions in terms of sine or cosine in the integrand. This will guide the choice of substitution.
  • Use the substitution to transform the integrand into a more workable form. This often involves differentiating your chosen substitution with respect to the variable.
  • Substitute both the expression and the variable of integration to completely change the variable.
In this example, the choice of \( u = 1 + 2 \sin 4\theta \)allows us to express every part of the original integral in terms of \( u \),simplifying the integration process.
Power integral
The power integral is an essential concept in calculus, referring to integrals involving polynomial expressions of the form \( x^n \),where \( n \)is a real number. It is a straightforward type of integral that uses the power rule:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]as long as \( n eq -1 \).
In the context of our exercise, once the substitution was made, the integral was reduced to a simpler form:\[ \int \frac{1}{8u^4} \, du = \frac{1}{8} \int u^{-4} \, du \]which is a classic power integral.
  • Recognize the form of the integrand so that the power rule can be applied.
  • The process involves adjusting the coefficient (in this case \( \frac{1}{8}\)already factored out) and applying the integral directly.
  • After integration, remember to simplify and address the constant \( C \).This step is important as integration results in an entire family of functions.
Executing this correctly allows us to find the antiderivative efficiently without the back-and-forth complex integrations.
Integral calculus
Integral calculus is a fundamental part of mathematical analysis focused on the accumulation of quantities, such as areas under curves. It broadly deals with integrals, both definite and indefinite, and their properties. In the exercise given, we tackled an indefinite integral, aiming to find a general function whose derivative gives us the original function.
  • The process begins with identifying the type of integral. In scenarios like ours, substitution simplifies the function, enabling easier manipulation.
  • Integral calculus often involves changing variables (as we did with trigonometric substitution) and applying rules like the power rule.
  • Once the simpler form is achieved, we integrate regarding the new variable, reverting to the original variable after simplification.
This integration gives us a function plus an additional constant \( C \),representing all possible vertical shifts of the antiderivative. Integral calculus is crucial in fields like physics, engineering, and beyond, helping us evaluate cumulative quantities.

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Most popular questions from this chapter

Let \(F(x)=\int_{4}^{x} \sqrt{t^{2}+9} d t .\) Find $$ \begin{array}{llll}{\text { (a) } F(4)} & {\text { (b) } F^{\prime}(4)} & {\text { (c) } F^{\prime \prime}(4)} & {\text { . }}\end{array} $$

A traffic engineer monitors the rate at which cars enter the main highway during the afternoon rush hour. From her data she estimates that between 4: 30 P.M. and 5: 30 P. M. the rate \(R(t)\) at which cars enter the highway is given by the formula \(R(t)=100\left(1-0.0001 t^{2}\right)\) cars per minute, where \(t\) is the time (in minutes) since 4: 30 P.M. (a) When does the peak traffic flow into the highway occur? (b) Estimate the number of cars that enter the highway during the rush hour.

Express \(F(x)\) in a piecewise form that does not involve an integral. $$ F(x)=\int_{-1}^{x}|t| d t $$

(a) Let $$I=\int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)} d x$$ Show that \(I=a / 2\). [Hint: Let \(u=a-x,\) and then note the difference between the resulting integrand and \(1 .]\) (b) Use the result of part (a) to find $$\int_{0}^{3} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} d x$$ (c) Use the result of part (a) to find $$\int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x$$

Let \(f\) denote a function that is continuous on an interval \([a, b],\) and let \(x^{*}\) denote the point guaranteed by the Mean-Value Theorem for Integrals. Explain geometrically why \(f\left(x^{*}\right)\) may be interpreted as "mean" or average value of \(f(x)\) over \([a, b] .\) (In Section 5.8 we will discuss the concept of "average value" in more detail.)
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