Calculus for Team-Based Inquiry Learning: 2022 Edition

Activity 5.4.1.

Consider the integral \(\displaystyle \int \frac{\sqrt{1-x^2}}{x} dx\text{.}\) If one were to use the method of substitution to evaluate this integral, what would be the best choice of \(u\text{?}\)

1. \(\displaystyle u=1-x^2\)

2. \(\displaystyle u=\sqrt{1-x^2}\)

3. \(\displaystyle u=\frac{1}{x}\)

4. There is no u for which the method of substitution will work.

Activity 5.4.2.

It would be good if \(\sqrt{1-x^2}\) were a term that can be written more simply. Which of the following mathematical facts could help us simplify this integrand?

1. For any real number, \(x^2\) must be positive (note: this isn’t even true).

2. \(\displaystyle u=\sqrt{1-x^2}\)

3. The Pythagorean theorem allows us to write \((\cos(\theta))^2\) as \(1-(\sin(\theta))^2\text{.}\)

4. The ratio of an even function and an odd function is odd.

Activity 5.4.3.

Suppose we let \(x=\sin(\theta)\text{.}\) Then \(\frac{dx}{d\theta}=\cos(\theta)\) and \(dx=\cos(\theta) d\theta\text{.}\) Which of the following best represents the integral \(\displaystyle \int \sqrt{1-x^2}dx\text{?}\)

1. \(\displaystyle \displaystyle \int \sqrt{1-\sin^2(\theta)}dx\)

2. \(\displaystyle \displaystyle \int \sqrt{ 1-\sin^2(\theta)\cos(\theta)}d\theta\)

3. \(\displaystyle \displaystyle \int \sqrt{ 1-\sin^2(\theta)}\cos(\theta)d\theta\)

Activity 5.4.4.

Use the fact from Activity 5.4.2 to rewrite your choice of integral from Activity 5.4.3 to an integral without any \(\sin(\theta)\) terms.

1. \(\displaystyle \displaystyle\int \cos(\theta) d\theta\)

2. \(\displaystyle \displaystyle\int \cos^2(\theta) d\theta\)

3. \(\displaystyle \displaystyle\int \sqrt{\cos(\theta)} d\theta\)

Activity 5.4.5.

Use the fact that \(\cos^2(\theta) =\frac{1+\cos(2\theta)}{2}\) to rewrite the integral from Activity 5.4.4.

1. \(\displaystyle \displaystyle\int \sqrt{\frac{1+\cos(2\theta)}{2}} d\theta\)

2. \(\displaystyle \displaystyle\int \frac{1+\cos(2\theta)}{2} d\theta\)

3. \(\displaystyle \displaystyle\int \left(\frac{1+\cos(2\theta)}{2} \right)^2d\theta\)

Activity 5.4.6.

Evaluate the integral in terms of \(\theta\) that you found in Activity 5.4.5.

Activity 5.4.7.

Consider the fact that \(x=\sin(\theta)\) and, thus, \(\theta=\arcsin(x)\text{.}\) Use this to rewrite the solution you found in Activity 5.4.6 in terms of \(x\text{.}\)

Activity 5.4.8.

Consider the integral \(\displaystyle \int \sqrt{9-x^2} dx\text{.}\) How can we best modify our approach in Activity 5.4.3 and adapt it here?

1. Let \(x=\sin(\theta)\text{.}\) It follows that \(dx=\cos(\theta)d\theta\) and, subsequently, \(\displaystyle \int \sqrt{9-x^2} dx= \int \sqrt{9-\sin^2(\theta)} \cos(\theta) d\theta\text{.}\)

2. Let \(x=3\sin(\theta)\text{.}\) It follows that \(dx=3\cos(\theta)d\theta\) and, subsequently, \(\displaystyle \int \sqrt{9-x^2} dx= \int \sqrt{9-(3\sin(\theta))^2} \cos(\theta) d\theta\text{.}\)

3. Let \(x=\sin(3\theta)\text{.}\) It follows that \(dx=3\cos(3\theta)d\theta\) and, subsequently, \(\displaystyle \int \sqrt{9-x^2} dx= \int \sqrt{9-\sin^2(3\theta)} 3\cos(3\theta) d\theta\text{.}\)

Activity 5.4.9.

Use your choice of substitution in Activity 5.4.8 and the fact that \(\cos^2(\theta) =\frac{1+\cos(2\theta)}{2}\) to rewrite the integral from Activity 5.4.8

1. \(\displaystyle \displaystyle \int \sqrt{\frac{9+9\cos(2\theta)}{2}} \cos(\theta) d\theta\)

2. \(\displaystyle \displaystyle \int \frac{9+9\cos(2\theta)}{2} \cos(\theta) d\theta\)

3. \(\displaystyle \displaystyle \int \left(\frac{9+9\cos(2\theta)}{2}\right)^2 \cos(\theta) d\theta\)

Activity 5.4.10.

Evaluate the integral that you found in Activity 5.4.9. (Recall that since \(x=3\sin(\theta), \theta=\arcsin\left(\frac{x}{3}\right)\text{.}\))

Activity 5.4.11.

Consider the integral \(\displaystyle \int \frac{1}{1+x^2} dx\text{.}\) Which of the following facts would best simplify this expression?

1. \(\displaystyle \cos^2(\theta) =\frac{1+\cos(2\theta)}{2}\)

2. \(\displaystyle (\cos(\theta))^2=1-(\sin(\theta))^2\)

3. Since \(\sin^2(\theta) + \cos^2(\theta)=1, \tan^2(\theta)+1=\sec^2(\theta)\text{.}\)

Activity 5.4.12.

Suppose we let \(x=\tan(\theta), dx =\sec^2(\theta)\text{.}\) Which of the following best represents the integral \(\displaystyle \int \frac{1}{1+x^2} dx\text{?}\)

1. \(\displaystyle \displaystyle \int \frac{1}{1+\tan^2(\theta)} dx\)

2. \(\displaystyle \displaystyle \int \frac{1}{1+\tan^2(\theta)\sec^2(\theta)} d\theta\)

3. \(\displaystyle \displaystyle \int \frac{1}{1+\tan^2(\theta)} \sec^2(\theta)d\theta\)

Activity 5.4.13.

Using the fact from Activity 5.4.11, simplify and evaluate your choice of integral from Activity 5.4.12. (Recall that since \(x=\tan(\theta), \theta=\arctan(x)\text{.}\))

Activity 5.4.14.

Consider the integral \(\displaystyle \int \frac{1}{\frac{1}{4}+x^2} dx\text{.}\) How can we best modify our approach in Activity 5.4.12 and adapt it here?

1. Let \(x=\frac{1}{2}\tan(\theta)\text{.}\) It follows that \(dx=\frac{1}{2}\sec^2(\theta)d\theta\) and, subsequently, \(\displaystyle \int \frac{1}{\frac{1}{4}+x^2} dx = \int \frac{1}{\frac{1}{4}+\left(\frac{1}{2}\tan(\theta) \right)^2} \frac{1}{2} \sec^2(\theta)d\theta\text{.}\)

2. Let \(x=\frac{1}{4}\tan(\theta)\text{.}\) It follows that \(dx=\frac{1}{4}\sec^2(\theta)d\theta\) and, subsequently, \(\displaystyle \int \frac{1}{\frac{1}{4}+x^2} dx = \int \frac{1}{\frac{1}{4}+\left(\frac{1}{4}\tan(\theta) \right)^2} \frac{1}{4}\sec^2(\theta) d\theta\text{.}\)

3. Let \(x=\tan\left(\frac{1}{2}\theta\right)\text{.}\) It follows that \(dx=\sec^2\left(\frac{1}{2}\theta\right)d\theta\) and, subsequently, \(\displaystyle \int \frac{1}{\frac{1}{4}+x^2} dx = \int \frac{1}{\frac{1}{4}+\left(\tan\left(\frac{1}{2}\theta\right) \right)^2} \frac{1}{2} \sec^2\left(\frac{1}{2}\theta\right)d\theta\text{.}\)

Activity 5.4.15.

Evaluate the choice of integral you found in Activity 5.4.14, using the fact that if \(x=\frac{1}{2}\tan(\theta), \theta=\arctan(2x)\text{.}\)

Activity 5.4.16.

Consider the integral \(\displaystyle \int \sqrt{x^2-1} dx\text{.}\) Which of the following facts best simplifies this integral?

1. \(\displaystyle \cos^2(\theta) =\frac{1+\cos(2\theta)}{2}\)

2. \(\displaystyle \tan^2(\theta)+1=\sec^2(\theta)\)

3. Since \(\sin^2(\theta) =1- \cos^2(\theta), \tan^2(\theta)=\sec^2(\theta)-1\text{.}\)

Activity 5.4.17.

Using your choice in Activity 5.4.16, pick an appropriate trignometric substitution for \(x\) and rewrite \(\displaystyle \int \sqrt{x^2-1} dx\) as an integral in terms of \(\theta\text{.}\) (Remember to compute \(dx\) as well!)

Activity 5.4.18.

Evaluate the integral you found in Activity 5.4.17, using the fact that if \(x=\sec(\theta), \theta=\arcsec(x)\text{.}\)

Hint.

To better understand the relationship between \(x\) and functions of \(\theta\text{,}\) consider this right triangle where \(\sec(\theta)=x\text{.}\)

Activity 5.4.19.

If we wish to simplify an expression that contained a \(k^2-x^2\) expression, which would be the best substitution?

1. \(\displaystyle x=k\sin(\theta)\)

2. \(\displaystyle x=k\tan(\theta)\)

3. \(\displaystyle x=k\sec(\theta)\)

Activity 5.4.20.

If we wish to simplify an expression that contained a \(k^2+x^2\) expression, which would be the best substitution?

1. \(\displaystyle x=k\sin(\theta)\)

2. \(\displaystyle x=k\tan(\theta)\)

3. \(\displaystyle x=k\sec(\theta)\)

Activity 5.4.21.

If we wish to simplify an expression that contained a \(x^2-k^2\) expression, which would be the best substitution?

1. \(\displaystyle x=k\sin(\theta)\)

2. \(\displaystyle x=k\tan(\theta)\)

3. \(\displaystyle x=k\sec(\theta)\)

Activity 5.4.22.

Consider the unit circle \(x^2+y^2=1\text{.}\) Find a function \(f(x)\) so that \(y=f(x)\) is the graph of the upper-half semicircle of the unit circle.

Activity 5.4.23.