/*! 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 40 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 40

Evaluate the following integrals. $$\int \frac{d \theta}{1-\csc \theta}$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int \frac{d\theta}{1-\csc{\theta}}\). Answer: \(\int \frac{d \theta}{1-\csc \theta} = \ln\left|\sin{\theta}-1 + \sqrt{-(\sin{\theta}-1)^2(\sin{\theta}+1)}\right| + C\)

Step by step solution

01

Recall the definition of cosecant function

The cosecant function, denoted by \(\csc{\theta}\), is defined as the reciprocal of the sine function, i.e. \(\csc{\theta} = \frac{1}{\sin{\theta}}\). We can use this definition to rewrite our integral as follows: $$\int \frac{d\theta}{1-\frac{1}{\sin{\theta}}}$$
02

Simplify the expression inside the integral

Apply common denominator to the expression inside the integral to get the following: $$\int \frac{d\theta}{\frac{\sin{\theta} - 1}{\sin{\theta}}}$$ Now, invert and multiply the expression inside the integral: $$\int \frac{\sin{\theta}}{\sin{\theta}-1} d\theta$$
03

Substitute \(u = \sin{\theta} - 1\)

Let \(u = \sin{\theta} - 1\). Then, \(\frac{du}{d\theta} = \cos{\theta}\) and \(d\theta = \frac{du}{\cos{\theta}}\). Since \(u = \sin{\theta} - 1\), we have \(\sin{\theta} = u+1\): $$\int \frac{u+1}{u} \frac{du}{\cos{\theta}}$$
04

Rewrite \(\cos{\theta}\) using \(u\)

Now, \(\sin^2{\theta} + \cos^2{\theta} = 1\). Since \(\sin{\theta} = u+1\), we have \(\cos^2{\theta} = 1 - (u+1)^2 = 1 - (u^2 + 2u + 1) = -u^2 - 2u\). Hence, \(\cos{\theta} = \pm\sqrt{-u(u+2)}\). Now the integral becomes: $$\int \frac{u+1}{u} \frac{du}{\pm\sqrt{-u(u+2)}}$$
05

Simplify and evaluate the integral

Simplifying the above integral: $$\int \frac{du}{\pm\sqrt{-u^2(u+2)}}$$ Join the \(\pm\) sign in the constant \(C\). $$\int \frac{du}{\sqrt{-u^2(u+2)}}$$ We can recognize it as an integral of the form \(\int \frac{du}{\sqrt{u^2-a^2}}\). The standard formula for this type of integral is given by: $$\int \frac{du}{\sqrt{u^2-a^2}} = \ln\left|u+\sqrt{u^2-a^2}\right| + C$$ Using this formula, we get: $$\int \frac{du}{\sqrt{-u^2(u+2)}} = \ln\left|u + \sqrt{-u^2(u+2)}\right| + C$$
06

Substitute back \(\sin{\theta}\) and simplify

Now, substitute back \(u = \sin{\theta}-1\): $$\int \frac{d\theta}{1-\csc \theta} = \ln\left|\sin{\theta}-1 + \sqrt{-(\sin{\theta}-1)^2(\sin{\theta}+1)}\right| + C$$ Thus, we have evaluated the integral: $$\int \frac{d \theta}{1-\csc \theta} = \ln\left|\sin{\theta}-1 + \sqrt{-(\sin{\theta}-1)^2(\sin{\theta}+1)}\right| + C$$

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Most popular questions from this chapter

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t)\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume that \(s\) is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=\cos a t \longrightarrow F(s)=\frac{s}{s^{2}+a^{2}}$$

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