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Chapter 6: Problem 25

Use integration tables to evaluate the integral. $$ \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+\sin ^{2} x} d x $$

Short Answer

Expert verified
The value of the integral is 0.

Step by step solution

01

Substitute variable

Let's start the solution by substituting \(\cos x = t\). This implies that \(-\sin x\, dx = dt\). After this substitution, the integral becomes \[\int \frac{t}{1 + (1 - t^{2})}(-dt)\].
02

Simplify the expression

In this step, simplify the obtained integral. When simplifying, you get \[\int \frac{-t dt}{2 - t^{2}}\].
03

Evaluate the integral

Evaluate the integral using the rule for the integral \(\int \frac{du}{a^{2} - u^{2}} = \frac{1}{a}\ln |\frac{a + u}{a - u}|\) for \(u = t\) and \(a = \sqrt{2}\). After applying this rule, the outcome becomes \[-\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2} + t}{\sqrt{2} - t}\right| + C\].
04

Substitute back the original variable

In this step, replace \(t\) with \(\cos x\). The result is \[-\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2} + \cos x}{\sqrt{2} - \cos x}\right| + C\].
05

Find the definite integral

Finally, find the value for the definite integral. Using the variables of integration we have \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). After applying these limits, the result is 0, because the function is an odd function and the limits are symmetrical around 0.

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Most popular questions from this chapter

In Exercises 65 and 66, apply the Extended Mean Value Theorem to the functions \(f\) and \(g\) on the given interval. Find all values \(c\) in the interval \((a, b)\) such that \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\) \(\begin{array}{l} \underline{\text { Functions }} \\ f(x)=\sin x, \quad g(x)=\cos x \end{array} \quad \frac{\text { Interval }}{\left[0, \frac{\pi}{2}\right]}\)

Prove that \(I_{n}=\left(\frac{n-1}{n+2}\right) I_{n-1},\) where \(I_{n}=\int_{0}^{\infty} \frac{x^{2 n-1}}{\left(x^{2}+1\right)^{n+3}} d x, \quad n \geq 1 .\) Then evaluate each integral. (a) \(\int_{0}^{\infty} \frac{x}{\left(x^{2}+1\right)^{4}} d x\) (b) \(\int_{0}^{\infty} \frac{x^{3}}{\left(x^{2}+1\right)^{5}} d x\) (c) \(\int_{0}^{\infty} \frac{x^{5}}{\left(x^{2}+1\right)^{6}} d x\)

Describe the different types of improper integrals

(b) Use the result of part (a) to find the equation of the path of the weight. Use a graphing utility to graph the path and compare it with the figure. (c) Find any vertical asymptotes of the graph in part (b). (d) When the person has reached the point (0,12) , how far has the weight moved?A person moves from the origin along the positive \(y\) -axis pulling a weight at the end of a 12 -meter rope (see figure). Initially, the weight is located at the point (12,0) . (a) Show that the slope of the tangent line of the path of the weight is $$ \frac{d y}{d x}=-\frac{\sqrt{144-x^{2}}}{x} $$

Prove the following generalization of the Mean Value Theorem. If \(f\) is twice differentiable on the closed interval \([a, b],\) then \(f(b)-f(a)=f^{\prime}(a)(b-a)-\int_{a}^{b} f^{\prime \prime}(t)(t-b) d t\).
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