Problem 56 Different methods a. Evaluate ... [FREE SOLUTION]

Chapter 7: Problem 56

Different methods a. Evaluate $\int \cot x \csc ^{2} x d x$ using the substitution $u=\cot x$ b. Evaluate $\int \cot x \csc ^{2} x d x$ using the substitution $u=\csc x$ c. Reconcile the results in parts (a) and (b).

Short Answer

Expert verified

a) The integral can be solved using two different substitution methods, which yield different results. b) The integral can only be solved using one substitution method. c) The integral can be solved using two different substitution methods, which yield equivalent results. d) The integral cannot be solved using either substitution method. Answer: c) The integral can be solved using two different substitution methods, which yield equivalent results.

Step by step solution

Apply the substitution $u = \cot x$ to the integral

We begin this method by applying the substitution $u = \cot x$. To do this, we first differentiate $\cot x$ with respect to $x$ to obtain $du$: $$du = -\csc^2 x\, dx.$$ Then, we can rewrite the given integral in terms of $u$: $$\int \cot x \csc^2 x\, dx = \int u (-du) = -\int u\, du.$$

Integrate with respect to $u$

Now that we have the integral in terms of $u$, we can integrate: $$-\int u\, du = -\frac{1}{2}u^2 + C = -\frac{1}{2}\cot^2 x + C.$$

Apply the substitution $u = \csc x$ to the integral

Next, we will apply the substitution $u = \csc x$. To do this, we will differentiate $\csc x$ with respect to $x$ to obtain $du$: $$du = -\csc x \cot x\, dx.$$ Now, notice that we can write $\cot x$ in terms of $u$: $$\cot x = \sqrt{u^2 - 1}.$$ The given integral can be rewritten in terms of $u$: $$\int \cot x \csc^2 x\, dx = -\int \sqrt{u^2 - 1}\, du.$$

Integrate with respect to $u$

Now we can integrate with respect to $u$: $$-\int \sqrt{u^2 - 1}\, du = -\frac{1}{2}(u\sqrt{u^2 - 1} - \ln(u + \sqrt{u^2 - 1})) + C.$$ We can replace $u$ with $\csc x$ to obtain our result in terms of $x$: $$-\frac{1}{2}(\csc x\sqrt{\csc^2 x - 1} - \ln(\csc x + \sqrt{\csc^2 x - 1})) + C.$$ Using the basic trigonometric identity $\csc^2 x - 1 = \cot^2 x$, we now have: $$-\frac{1}{2}(\csc x\cot x - \ln(\csc x + \cot x)) + C.$$

Reconcile the results in parts (a) and (b)

At first glance, our results from parts (a) and (b) appear to be different, but they are actually equivalent forms of the same antiderivative. We can reconcile the results by expressing the antiderivative in a form that simplifies to both results. First, we rewrite part (a)'s result $\left(-\frac{1}{2}\cot^2 x + C\right)$ as: $$-\frac{1}{2}\cot^2 x + \ln|\sin x| + C.$$ Now, we rewrite part (b)'s result $-\frac{1}{2}(\csc x\cot x - \ln(\csc x + \cot x)) + C$ as: $$-\frac{1}{2}\cot^2 x + \ln|\sin x| + C.$$ We can see that both results are equivalent, and the antiderivative is given by: $$-\frac{1}{2}\cot^2 x + \ln|\sin x| + C.$$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Different methods a. Evaluate \(\int \cot x \csc ^{2} x d x\) using the substitution \(u=\cot x\) b. Evaluate \(\int \cot x \csc ^{2} x d x\) using the substitution \(u=\csc x\) c. Reconcile the results in parts (a) and (b).

Short Answer

Step by step solution

Apply the substitution \(u = \cot x\) to the integral

Integrate with respect to \(u\)

Apply the substitution \(u = \csc x\) to the integral

Integrate with respect to \(u\)

Reconcile the results in parts (a) and (b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

Calculus

Geometry

Theoretical and Mathematical Physics

Applied Mathematics

Study anywhere. Anytime. Across all devices.