I think this is the millionth time that I’ve started a blog by saying this, but I had another tough week this week and didn’t make it very far through KA. As usual, it wasn’t due to a lack of effort! I wrote out 36 pages of notes but a lot of my time was spent reviewing trigonometric identities which resulted in me not making it through the final two sections of Integrals. One thing I was happy about this week, however, was that I found the algebra Sal used in the trig identity proofs very straightforward and simple to comprehend. I remember struggling to understand these proofs when I first learned them, feeling completely overwhelmed and like they were way out of my depth. It was a great feeling working through them this week and having no issue following along. Woo!! 🥳

I picked up this week working on the second and final exercise of the section Integrating Using Long Division and Completing the Square. This exercise was all about using the ‘completing the square’ method to simplify polynomials inside the integrand in order to evaluate them. It first took me a bit of review to even remember what completing the square was and how to do it. After I figured it out, what I then found much harder was reviewing the derivatives of arctan and arcsine, a.k.a. tan^-1(x) and sin^-1(x) respectively. The reason I had to review these derivatives was because the exercise had the derivatives of both inverse functions inside the integrand and so I needed to understand them, in general, and be able to recognize them. I watched this video which helped me figure it out:

• y = tan^-1(x)
• tan(y) = tan(tan^-1(x))
  • (Note: On the right side of the equation, tan cancels out tan^-1 since they’re inverse operations. It’s like multiplying and dividing any number by another number.)
• tan(y) = x
• d/dx[tan(y)] = d/dx [x]
  • a(x) = tan(x)
  • a’(x) = sec²(x)
    • (Note: I’m not going to show the derivation of d/dx[tan(x)] = sec²(x), you’re just gonna have to take my word for it.)
  • b(x) = y
  • b’(x) = dy/dx
• d/dx[a(b(x))] = d/dx[x]
• a’(b(x)) * b’(x) = 1
• sec²(y) * dy/dx = 1
• dy/dx = 1/(sec²(y))
  • (Note: At this point, you need to use the trig identity sin²(x) + cos²(x) = 1 to turn sec²(y) into tan²(y).)
    • sin²(x) + cos²(x) = 1
    • sin²(x)/cos²(x) + cos²(x)/cos²(x) = 1/cos²(x)
    • tan²(x) + 1 = sec²(x)
• dy/dx = 1/(tan²(x) + 1)
  • (Note: From the very top, you can see that tan(y) = x so you can square both sides of the equation in order to replace tan²(y) with x².)
  • tan(y) = x
  • (tan(y))² = x² = tan²(y)
• dy/dx = 1/(x² + 1)

Once I reviewed this video and got a better feel for the derivatives of arctan and arcsine, I was then able to get through the exercise pretty handily. Here’s an example question I worked through:

• ∫1/(5x² – 30x + 65) * dx
  • = ∫1/5 * 1/(x² – 6x + 13) * dx
  • = 1/5 * ∫1/(x² – 6x + 9 – 9 + 13) * dx
  • = 1/5 * ∫1/((x² – 6x + 9) – 9 + 13) * dx
  • = 1/5 * ∫1/((x² – 6x + 9) + 4) * dx
  • = 1/5 * ∫1/((x – 3)² + 4) * dx
    • (Note: At this point, you need to make the integrand take the shape of the derivative of arctan, 1/x² + 1, so you need to factor out a 4 in order to make the 4 turn into a 1. You do this by dividing both terms in the denominator by 2² which allows you to turn (x – 3)²/2² into ((x – 3)/2)² which you can then state is u (what I think of as b(x)) when using u-substitution.)
  • = 1/5 * ∫1/(((x – 3)² + 4)/2²) * dx
  • = 1/5 * ∫1/1/2² * 1/(((x – 3)/2)² + 1) * dx
  • = 1/5/2² * ∫1/(((x – 3)/2)² + 1) * dx
  • = 1/20 * ∫1/(((x – 3)/2)² + 1) * dx
    • A(x) = tan^-1(x)
    • a’(x) = 1/(x² + 1)
    • b(x) = (x – 3)/2
      • = x/2 – 3/2
    • b’(x) = ½
  • = 1/20 * ∫a’(b(x)) * 2/2 * dx
  • = 1/20 * 2 * ∫a’(b(x)) * 1/2 * dx
  • = 1/10 * ∫a’(b(x)) * b’(x) * dx
  • = 1/10 * (A(b(x))) + C
  • = 1/10 * tan^-1((x – 3)/2) + C

After finishing off that exercise, I then moved into the final section of Integrals which was titled Integrating Using Trigonometric Identities. It proved to be hella difficult. For context, I spent Wednesday, Thursday, Friday, and Saturday watching 3 videos and attempting 1 exercise which I failed miserably. I had no intuition for Sal’s work in the videos and went 0/6 when trying the questions in the exercise. The following are 3 example questions I attempted and how to solve each of them. For the sake of time, I’m only going to write out the math for the first 2 questions.

            Question 1:

• ∫cos³(x) * dx
  • = ∫cos²(x)cos(x) * dx
    • (Note: cos²(x) can be rewritten as 1 – sin²(x) using the trig identity cos²(x) + sin²(x) = 1.)
  • = ∫(1 – sin²(x))cos(x) * dx
    • (Note: At this step you expand the integrand.)
  • = ∫cos(x) – sin²(x)cos(x) * dx
  • = ∫cos(x) * dx – ∫sin²(x)cos(x) * dx
    • A(x) = x³/3
    • a'(x) = x²
    • b(x) = sin(x)
    • b’(x) = cos(x)
  • = ∫cos(x) * dx – ∫a’(b(x))b’(x) * dx
  • = –sin(x) – a(b(x)) + C
  • = –sin(x) – sin³(x)/3

Question 2:

(Note: This question really bugged me because it told me that I essentially needed to memorize the answer if I was going to be able to solve it. I don’t like the idea of simply memorizing math concepts and not intuitively understanding what’s going on.)

• ∫sec(x) * dx
  • = ∫sec(x) * ((sec(x) + tan(x))/(sec(x) + tan(x)) * dx
  • = ∫(sec²(x) + sec(x)tan(x))/(sec(x) + tan(x)) * dx
  • = ∫(sec²(x) + sec(x)tan(x)) * (sec(x) + tan(x))^-1 * dx
  • = ∫(sec²(x) * (sec(x) + tan(x))^-1 + sec(x)tan(x)) * (sec(x) + tan(x))^-1 * dx
    • A(x) = sec(x)
    • a'(x) = sec(x)tan(x)
    • b(x) = tan(x)
    • b’(x) = sec²(x)
    • c(x) = ln(|x|)
    • c’(x) = 1/x = x^-1
  • = ∫b’(x) * c’(a(x) + b(x)) + a’(x) * c’(a(x) + b(x)) * dx
    • (Note: Here you need to factor out the c’(a(x) + b(x)) from both terms.)
  • = ∫(c’(a(x) + b(x)))(b’(x) + a’(x)) * dx
  • = c(a(x) + b(x)) + C
  • = ln(|sec(x) + tan(x)|)

Question 3:

As you can see from my little addendum on my second page, using my a(x), b(x) method for this question got a little out of control and it seemed like using u and du/dx might have been the better way to go. I redid the question and tried using proper u-substitution and it did seem a bit easier.

To wrap this post up on a more positive note – I completed the Algebra 2 Course Challenge this week for SchoolHouse! The only problem was that the video didn’t upload properly so I had to download it and email it to them. I did that on Friday and haven’t heard back from them yet so I’m hoping it all works out and I get the certification to tutor algebra 2. 👍🏻

As I thought it might, it looks like it’s going to come down to the wire in terms of me getting through this unit, Integrals (2,480/3,200 M.P.), before the start of the New Year. I’m so close but I think that finishing up this exercise on integrating with trig identities might take me at least a few days. I also have a feeling I’m going to need to do the unit test at least a few times before I pass it. I’ll do my best to get through it all but, as I’ve said many times before, I’m more concerned with learning the material than hitting milestones along a certain timeline. As long as I’m putting in a minimum of 5 hours a week and a good effort overall, I don’t care too much if I don’t get through this unit by the end of the year.