/*! 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 36 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 36

Evaluate the following integrals. $$\int \frac{\sec ^{2} x}{\tan ^{5} x} d x$$

Short Answer

Expert verified
Based on the above step by step solution, the short answer to the problem \(\int \frac{\sec^2 x}{\tan^5 x} dx\) is \(-\frac{1}{4\tan^4 x} + C\).

Step by step solution

01

Identify the substitution

We will substitute \(u = \tan x\). The reason is that the derivative of \(\tan x\) gives us \(\sec^2 x\) which is present in the integral. This substitution will help us eliminate \(\sec^2 x\) and leave us with an integral in terms of \(u\).
02

Find the derivative of the substitution

Now let's find the derivative of our substitution. \(u = \tan x\), so we need to find \(\frac{du}{dx}\): \[ \frac{du}{dx} = \frac{d}{dx} \tan x = \sec^2 x \]
03

Rewrite the integral in terms of u

Now we will rewrite the integral in terms of \(u\). Since \(u = \tan x\), we know that \(u^5 = \tan^5 x\). We will replace \(\tan^5 x\) with \(u^5\). Also, we have \(\frac{du}{dx} = \sec^2 x\), so we can replace \(dx\) with \(du\) by multiplying both sides by \(dx\) and dividing by \(\sec^2 x\). We get \(\frac{du}{\sec^2 x} = dx\). So, the new integral will be: \[ \int \frac{\sec^2 x}{\tan^5 x} dx = \int \frac{1}{u^5} du \]
04

Evaluate the integral

Now we can evaluate the integral in terms of \(u\): \[ \int \frac{1}{u^5} du = -\frac{1}{4u^4} + C \]
05

Substitute tan x back into the result

Since we have found the antiderivative in terms of \(u\), now we need to substitute back \(\tan x\) for \(u\): \[ -\frac{1}{4u^4} + C = -\frac{1}{4\tan^4 x} + C \] So, the final solution for the given integral is: \[ \int \frac{\sec^2 x}{\tan^5 x} dx = -\frac{1}{4\tan^4 x} + C \]

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int \frac{4 x^{6}}{x^{4}+3 x^{2}} d x\), the first step is to find the partial fraction decomposition of the integrand. b. The easiest way to evaluate \(\int \frac{6 x+1}{3 x^{2}+x} d x\) is with a partial fraction decomposition of the integrand. c. The rational function \(f(x)=\frac{1}{x^{2}-13 x+42}\) has an irreducible quadratic denominator. d. The rational function \(f(x)=\frac{1}{x^{2}-13 x+43}\) has an irreducible quadratic denominator.

Suppose that a function \(f\) has derivatives of all orders near \(x=0 .\) By the Fundamental Theorem of Calculus, \(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\) a. Evaluate the integral using integration by parts to show that $$f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t.$$ b. Show (by observing a pattern or using induction) that integrating by parts \(n\) times gives $$\begin{aligned} f(x)=& f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0) \\ &+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots \end{aligned}$$ This expression is called the Taylor series for \(f\) at \(x=0\).

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{2+\cos x}$$

Refer to Theorem 2 and let \(f(x)=\sin e^{x}\) a. Find a Trapezoid Rule approximation to \(\int_{0}^{1} \sin \left(e^{x}\right) d x\) using \(n=40\) subintervals. b. Calculate \(f^{\prime \prime}(x)\) c. Explain why \(\left|f^{\prime \prime}(x)\right|<6\) on \([0,1],\) given that \(e<3\). (Hint: Graph \(\left.f^{\prime \prime} .\right)\) d. Find an upper bound on the absolute error in the estimate found in part (a) using Theorem 2.

By reduction formula 4 in Section 3 $$\int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\left(9-x^{2}\right)^{-2},\left[0, \frac{3}{2}\right]$$
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