How do you solve this calculus 2 integration problem?

In summary, there are two forms of the anti-derivative for the given integral, and they differ by a constant. This is justified by taking the derivative of both forms and obtaining the original integrand. Therefore, both forms are correct.
  • #1
harpazo
208
16
I am currently studying calculus 3. However, once in a while, I like to review single variable calculus. I decided to tackle the following calculus 2 integration problem:

∫ tan^3 x dx

Solution:

Separate into two tangent functions.

(tan^2 x)(tan x)

Rewrite tan^2 x as sec^2 x - 1 using its equivalent trig identity.

∫ (tan x)(sec^2 x - 1) dx

Let u = sec^2 x

du/dx = tan x

dx = du/ tan x

∫ (tan x)(u^2 - 1) du/tan x

∫ (u^2 - 1) du

(u^3/3) - u + C

Back-substitute for u.

We know that u = sec^2 x.

[(sec^2 x)^3]/3 - sec^2 x + C

What do you say?
 
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  • #2
If you differentiate your result w.r.t $x$, do you get your original integrand as the result?
 
  • #3
MarkFL said:
If you differentiate your result w.r.t $x$, do you get your original integrand as the result?

No. I did not get back the original problem. Can you go through my reply to see where I went wrong?
 
  • #4
Harpazo said:
No. I did not get back the original problem. Can you go through my reply to see where I went wrong?

We are given:

\(\displaystyle I=\int\tan^3(x)\,dx\)

I agree that a good strategy is to write this as:

\(\displaystyle I=\int\tan^2(x)\cdot\tan(x)\,dx\)

Now we can use the Pythagorean identity:

\(\displaystyle \tan^2(\theta)=\sec^2(\theta)-1\)

To write the integral as:

\(\displaystyle I=\int\tan(x)\left(\sec^2(x)-1\right)\,dx\)

Distribute:

\(\displaystyle I=\int\tan(x)\sec^2(x)-\tan(x)\,dx=\int \tan(x)\,\sec^2(x)\,dx+\int \frac{-\sin(x)}{\cos(x)}\,dx\)

Can you now spot the appropriate substitutions to be used on the two integrals?
 
  • #5
∫ (tan x)(sec^2 x) dx + ∫ -(sin x)/(cos x) dx

For the left side integral, let u = tan x

du/dx = sec^2 x

dx = du/sec^2 x

∫ u• sec^2 x • du/sec^2 x

∫ u du + ∫ -(sin x)/(cos x) dx

For the integral on the right side,
let u = cos x

du/dx = - sin x

dx = du/-sin x

∫ u du + ∫ -(sin x)/u • du/-sin x

∫ u du + ∫ 1/u • du

(u^2)/2 + ln u + C

Back-substitute for u.

[(sec^2 x)^2]/2 + ln (cos x) + C
 
  • #6
Harpazo said:
∫ (tan x)(sec^2 x) dx + ∫ -(sin x)/(cos x) dx

For the left side integral, let u = tan x

du/dx = sec^2 x

dx = du/sec^2 x

∫ u• sec^2 x • du/sec^2 x

∫ u du + ∫ -(sin x)/(cos x) dx

For the integral on the right side,
let u = cos x

du/dx = - sin x

dx = du/-sin x

∫ u du + ∫ -(sin x)/u • du/-sin x

∫ u du + ∫ 1/u • du

(u^2)/2 + ln u + C

Back-substitute for u.

[(sec^2 x)^2]/2 + ln (cos x) + C

For the first integral, you let:

\(\displaystyle u=\tan(x)\)

And so you would have:

\(\displaystyle I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C\)

Now, suppose that for the first integral, we let:

\(\displaystyle u=\sec(x)\implies du=\sec(x)\tan(x)\,dx\)

And so we eventually get:

\(\displaystyle I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C\)

How do we justify that both are correct? (Thinking)
 
  • #7
MarkFL said:
For the first integral, you let:

\(\displaystyle u=\tan(x)\)

And so you would have:

\(\displaystyle I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C\)

Now, suppose that for the first integral, we let:

\(\displaystyle u=\sec(x)\implies du=\sec(x)\tan(x)\,dx\)

And so we eventually get:

\(\displaystyle I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C\)

How do we justify that both are correct? (Thinking)

We take the derivative to check our answer.
 
  • #8
Harpazo said:
We take the derivative to check our answer.

Well, yes differentiating both forms gets us back to the original integrand, but how do we "justify" that both are correct?
 
  • #9
MarkFL said:
Well, yes differentiating both forms gets us back to the original integrand, but how do we "justify" that both are correct?

I do not understand your question.
 
  • #10
Harpazo said:
I do not understand your question.

In post #6, I gave two different anti-derivatives...how can we show both are correct?
 
  • #11
MarkFL said:
In post #6, I gave two different anti-derivatives...how can we show both are correct?

We can show they are both correct by setting them equal to each other.
 
  • #12
MarkFL said:
For the first integral, you let:

\(\displaystyle u=\tan(x)\)

And so you would have:

\(\displaystyle I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C\)

Now, suppose that for the first integral, we let:

\(\displaystyle u=\sec(x)\implies du=\sec(x)\tan(x)\,dx\)

And so we eventually get:

\(\displaystyle I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C\)

How do we justify that both are correct? (Thinking)

Would showing that the two anti-derivative forms differ by a constant suffice as justification?

\(\displaystyle I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C\)

\(\displaystyle I=\frac{1}{2}[\sec^2(x)-1]+\ln|\cos(x)|+C\)

\(\displaystyle I=\frac{1}{2}\sec^2(x)-\frac{1}{2}+\ln|\cos(x)|+C\)

\(\displaystyle I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C_2\)
 
  • #13
JorgeLuna said:
Would showing that the two anti-derivative forms differ by a constant suffice as justification?

\(\displaystyle I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C\)

\(\displaystyle I=\frac{1}{2}[\sec^2(x)-1]+\ln|\cos(x)|+C\)

\(\displaystyle I=\frac{1}{2}\sec^2(x)-\frac{1}{2}+\ln|\cos(x)|+C\)

\(\displaystyle I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C_2\)

Yes, that was exactly the kind of thing I was hoping the OP would state. (Yes)
 
  • #14
I don't know...
 

FAQ: How do you solve this calculus 2 integration problem?

What is Calculus 2?

Calculus 2 is the second course in a series of math courses that focuses on the study of change and motion. It includes topics such as integration techniques, applications of integrals, sequences and series, and parametric and polar equations.

Why is Calculus 2 important?

Calculus 2 builds upon the foundational concepts learned in Calculus 1 and provides a deeper understanding of mathematical functions and their applications. It is also a prerequisite for many advanced math and science courses, making it an important subject for students pursuing careers in these fields.

What are some common challenges in Calculus 2?

Some common challenges in Calculus 2 include grasping the concept of integration, mastering various integration techniques, and understanding the applications of integrals in real-life problems. Many students also struggle with the mathematical notation and symbols used in this course.

How can I succeed in Calculus 2?

To succeed in Calculus 2, it is important to have a strong foundation in Calculus 1 and to stay on top of the material by attending lectures, completing practice problems, and seeking help from professors or tutors when needed. It is also helpful to actively engage with the material by visualizing and understanding the concepts rather than just memorizing formulas.

How can I apply Calculus 2 in real life?

Calculus 2 has many real-life applications, such as calculating volumes of irregular shapes, finding the area under a curve to determine displacement or work done, and analyzing motion of objects. It is also used in fields such as engineering, physics, economics, and statistics to solve complex problems and make predictions based on mathematical models.

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