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Chapter 4: Problem 45

Determine the following indefinite integrals. Check your work by differentiation. $$\int \sec 4 \theta \tan 4 \theta d \theta$$

Short Answer

Expert verified
Question: Find the indefinite integral of the given function and check your work by differentiation: $$\int \sec 4 \theta \tan 4 \theta d \theta$$ Answer: The indefinite integral of the given function is $$\frac{1}{4} \ln|\sec 4\theta| + C$$

Step by step solution

01

Identify substitution

We want to find the indefinite integral: $$\int \sec 4 \theta \tan 4 \theta d \theta$$ To do so, we see that the product of secant and tangent is involved, so we will use the substitution \(u = 4\theta\) and \(du = 4 d\theta\). We rewrite the integral in terms of \(u\): $$\int \sec u \tan u \frac{1}{4} du$$
02

Perform the integration

Now we have the following integral: $$\frac{1}{4}\int \sec u \tan u du$$ Recall that the derivative of \(\ln |\sec u|\) is equal to \(\sec u \tan u\): $$\frac{d}{du}\ln|\sec u| = \sec u \tan u$$ So the integral becomes: $$\frac{1}{4} \ln|\sec u| + C$$
03

Substitute back

Now we need to rewrite the result in terms of \(\theta\). Recall that \(u = 4\theta\). Thus: $$\frac{1}{4} \ln|\sec 4\theta| + C$$ This is our final answer for the indefinite integral.
04

Check by differentiation

To check our result, we will differentiate the indefinite integral we found with respect to \(\theta\): $$\frac{d}{d\theta}\left(\frac{1}{4} \ln|\sec 4\theta| + C\right)$$ Using the chain rule: $$\frac{d}{d\theta}\left(\frac{1}{4} \ln|\sec 4\theta|\right)=\frac{1}{4}\left(\frac{1}{\sec 4\theta}\cdot\sec 4\theta \tan 4\theta \cdot 4\right)$$ Simplify this expression: $$\sec 4\theta \tan 4\theta$$ We see that the differentiated result is the same as the original function we wanted to integrate, which confirms that our result for the indefinite integral is correct.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. \(F(x)=x^{3}-4 x+100\) and \(G(x)=x^{3}-4 x-100\) are antiderivatives of the same function. b. If \(F^{\prime}(x)=f(x),\) then \(f\) is an antiderivative of \(F\) c. If \(F^{\prime}(x)=f(x),\) then \(\int f(x) d x=F(x)+C\) d. \(f(x)=x^{3}+3\) and \(g(x)=x^{3}-4\) are derivatives of the same function. e. If \(F^{\prime}(x)=G^{\prime}(x),\) then \(F(x)=G(x)\)

The functions \(f(x)=a x^{2},\) where \(a>0\) are concave up for all \(x\). Graph these functions for \(a=1,5,\) and 10, and discuss how the concavity varies with \(a\). How does \(a\) change the appearance of the graph?

Locate the critical points of the following functions and use the Second Derivative Test to determine whether they correspond to local maxima, local minima, or neither. $$f(x)=x^{3}+2 x^{2}+4 x-1$$

Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation \(a(t)=v^{\prime}(t)=g,\) where \(g=-9.8 \mathrm{m} / \mathrm{s}^{2}\). a. Find the velocity of the object for all relevant times. b. Find the position of the object for all relevant times. c. Find the time when the object reaches its highest point. What is the height? d. Find the time when the object strikes the ground. A stone is thrown vertically upward with a velocity of \(30 \mathrm{m} / \mathrm{s}\) from the edge of a cliff 200 m above a river.

Differentials Consider the following functions and express the relationship between a small change in \(x\) and the corresponding change in \(y\) in the form \(d y=f^{\prime}(x) d x\) $$f(x)=1 / x^{3}$$
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