/*! 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 39 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 39

Evaluate the following integrals. $$\int \frac{d x}{\sec x-1}$$

Short Answer

Expert verified
Question: Evaluate the integral of the function \(\frac{1}{\sec x - 1}\). Answer: The integral of the function \(\frac{1}{\sec x - 1}\) is: $$\int \frac{d x}{\sec x-1} = -\frac{1}{\sin \frac{x}{2}} + 2\sin \frac{x}{2} + C$$

Step by step solution

01

Rewrite the Integral using Trigonometric Identity

We need to rewrite the integral using a trigonometric identity. We know that \(\sec x = \frac{1}{\cos x}\), so we have the integral: $$\int \frac{d x}{\sec x-1} = \int \frac{d x}{\frac{1}{\cos x} - 1}$$ Let's simplify by multiplying both numerator and denominator by \(\cos x\): $$\int \frac{d x}{\frac{1}{\cos x} - 1} = \int \frac{\cos x \, d x}{1 - \cos x}$$ Now, we almost have a trigonometric expression that we can integrate. We will use the identity \(1 - \cos 2 \theta = 2 \sin^2 \theta\) to further simplify our expression.
02

Modify Trigonometric Identity and Use Substitution

In order to use the trigonometric identity \(1 - \cos 2 \theta = 2 \sin^2 \theta\), let's replace \(\theta\) with \(\frac{x}{2}\): $$1 - \cos x = 2 \sin^2 \frac{x}{2}$$ Now, our integral becomes: $$\int \frac{\cos x \, d x}{1 - \cos x} = \int \frac{\cos x \, d x}{2 \sin^2 \frac{x}{2}}$$ Let's use the substitution \(u = \sin \frac{x}{2}\). Therefore, \(d u = \frac{1}{2} \cos \frac{x}{2} d x\). Notice that \(\cos x = 1 - 2 \sin^2 \frac{x}{2} = 1-2u^2\). Substituting these expressions, we get: $$\int \frac{\cos x \, d x}{2 \sin^2 \frac{x}{2}} = \int \frac{(1 - 2u^2) d(2u)}{2 u^2} = \int \frac{1 - 2u^2}{u^2} d u$$
03

Evaluate the Integral

Now, we have an integral which we can easily solve: $$\int \frac{1 - 2u^2}{u^2} d u = \int \left(\frac{1}{u^2} - 2\right) d u$$ This integral can be solved by breaking it into separate parts: $$\int \left(\frac{1}{u^2} - 2\right) d u = \int \frac{1}{u^2} d u - \int 2 d u$$ Let's solve each part separately: $$\int \frac{1}{u^2} d u = -\frac{1}{u} + C_1$$ $$\int 2 d u = 2u + C_2$$ Therefore: $$\int \left(\frac{1}{u^2} - 2\right) d u = -\frac{1}{u} + 2u + C$$ where \(C = C_1 + C_2\).
04

Re-substitute the Variable

Now, we have to substitute \(x\) back into our solution. Recall that \(u = \sin \frac{x}{2}\), so: $$\int \left(\frac{1}{u^2} - 2\right) d u = -\frac{1}{\sin \frac{x}{2}} + 2\sin \frac{x}{2} + C$$ This is our final answer: $$\int \frac{d x}{\sec x-1} = -\frac{1}{\sin \frac{x}{2}} + 2\sin \frac{x}{2} + C$$

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Most popular questions from this chapter

Many methods needed Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x.\) b. Change variables by letting \(y=1 / x.\) c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) and conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0.\) d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\) (Source: Mathematics Magazine 59, No. 1 (February 1986): 49).

Evaluate the following integrals or state that they diverge. $$\int_{0}^{\pi / 2} \sec \theta d \theta$$

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{d x}{1+e^{x}}$$

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{2}{x\left(x^{2}+1\right)^{2}} d x$$

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 3 x \sin 2 x d x$$
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