In the ~2.5 years I’ve been working on KA, I’ve never had as devastating of a start to a week as I did this week. 😔 After having spent the last month reviewing the FToC and redoing the 5 exercises necessary to get my score back up to 80% M.P. (which is what you need before you start the unit test if you want to get 100% on the unit, overall), I got the very first question wrong on the unit test and just about had a nervous breakdown. The worst part was that the question I got wrong was EXACTLY the same kind of question I had been working on literally the day before. As much as I was ready to give up on math completely, I kept going on the test and finished the week getting through 28 of 32 questions and getting none wrong after the first one. That said, I cheated by looking up 2 or 3 of questions too confirm my answer but for the most part I’ve known how to solve all the questions on my own. Overall, I’m happy with how this week’s gone even though I clearly still haven’t completely figured out integrals.

I began the week working through the exercise Integration Using Completing the Square. I went through about 10 pages of notes on Tuesday working through 6 questions from this exercise and ended up getting all 6 of them correct on my first try. Even still, I ended the exercise still feeling like I had only a tenuous grasp on the antiderivatives that were used, (1/k)*arctan((1 + h)/k) and arcsin((1 + h)/k). I more-or-less memorized the formulas for each derivative and still don’t have an intuitive feel for how/why they work when working through these types of questions. Here’s one of the questions from my notes:

One of the most frustrating parts from this question is that the answer KA gives essentially states that I was somehow supposed to assume that I needed to figure out d/dx[(1/k) * arctan((x + h)/k)] in order to realize that it was the correct antiderivative to use to solve this question. As you can see from my notes, I can do the algebra to solve the derivative (although I find it a bit tricky) but I would never have known that the function in the integrand had the shape of the derivative of (1/k) * arctan((x + h)/k) if I hadn’t been told by KA. As I said, I ended up passing this exercise not because I fully understood the derivatives/antiderivatives of (1/k) * arctan((x + h)/k) or arcsin((x + h)/k) but simply because I memorized them. I’m also at a loss for why arccos(x) wasn’t used… 🤔

I started the unit test on Wednesday and, like I said, got the very first question wrong. The question asked me to find the antiderivative of ∫ 1/√(–x² –4x + 21) dx which I instantly recognized as the derivative of arcsin((x + h)/k), however I mixed up the formulas for arcsin((x + h)/k) and arctan((x + h)/k) and thought there was a 1/k in front of arcsin((x + h)/k). This is true for arctan((x + h)/k) but not for arcsin((x + h)/k). Here are the notes I took answering this question. You can see at the bottom of the page where I screwed up:

After getting this question wrong, I literally stared at my computer for 5 minutes and was ready to start crying… To make matters worse, I then couldn’t figure out the second question either! Question 2 asked me to find the antiderivative of ∫tan²(x)sec²(x) dx. I spent about 40 minutes trying to use trig-substitutions on it before giving up. I looked it up and realized that it was a u-sub question. Once I realized that, I solved the question in about 60 seconds. Here’s how it works:

• ∫tan²(x)sec²(x) dx = ∫(tan(x))² * sec²(x) dx
  • = ∫ (a’(b(x)) * b’(x)
    • a(x) = x³/3
    • a’(x) = x²
    • b(x) = tan(x)
    • b’(x) = sec²(x)
  • = a(b(x))
  • = tan³(x)/3 + C

As frustrated as I was that I didn’t figure out how to solve this question on my own, I was happy that I was able to quickly able solve it once I realized it was a u-sub question. 

From there, I made it all the way to the 17^th question before having any difficulty. On the 17^th question I was given the following information:

• F(x) = _-5∫^3x √(15 – t) dt
• F’(x) = ?

I vaguely remembered these types of questions but couldn’t remember exactly how to solve them. I was pretty sure that to solve for F’(x) all I needed to do was take the function inside the integrand, √(15 – t), and replace t with 3x and then also add a 3 in front of the function because the derivative of 3x is 3. As I explain below, this is a version of the chain rule but I couldn’t remember how it worked in the moment. I ended up being correct but I then went back and reviewed these types of questions to better understand what was going on. Here’s how to think about these types of questions:

• F(x) = _-5∫^3x √(15 – t) dt
  • F(x) = ₅∫^b(x) √(15 – t) dt
    • b(x) = 3x
    • b’(x) = 3
  • F(x) = a(b(x))
    • a(x) = ₅∫^x √(15 – t) dt
    • a’(x) = √(15 – x)
• F’(x) = a’(b(x)) * b’(x)
  • = √(15 – 3x) * 3

After making it through question 17, the next question I struggled with was question 22. This question asked me to find the antiderivative of ∫(x³ + x)/(x + 1) dx and after about 40 minutes of trying to use u-sub and thinking if there were any trig-subs I could use, I finally gave up and Googled it. I clicked on the first video that came up and within 5 seconds of watching the video I realized I needed to use polynomial long division on the function. It was frustrating that I didn’t realize this on my own but once I knew I needed to use this method, I was able to solve it on my own. Here’s my work:

In my notes you can see that Step 1 was me using polynomial long division on the function. I then inputted the quotient back into the integrand in Step 2.  From there, being that the quotient contained 4 separate terms, it was relatively easy to find the antiderivative of each term with the most difficult one being the final term, -2/(x + 1), which, as you can see, ends up as -2ln|x + 1|.

The final question I worked on this week, question 28, I also struggled with. The question asked me to find the value of the integral ₀∫^π/3 6tan(x)sec³(x)dx. Somewhat surprisingly, I managed to figure out that I needed to split sec³(x) into sec(x) and (sec(x))² and then use u-sub where u = sec(x) and du = tan(x)sec(x) (although, as usual, I used b(x) and b’(x) in place of u and du), but I then messed up the trigonometry. I double checked my answer on Symbolab before inputting it so I would have gotten the question wrong if I hadn’t double checked. Here’s the question and my work with the revised trig:

(INSERT SCREEN SHOT)

It’s not shown in my notes above, but the trigonometry I got wrong was thinking that cos(π/3) was equal to √3/2 but it’s equal to ½. Once I made that correction, I was able to solve it and get the correct answer.

I only have 4 questions left to get through at the beginning of this upcoming week. I’m very disappointed that I got the first question wrong this week and am also disappointed that I had to look up a few questions, but if I manage to get the remaining 4 questions correct, I’m planning to move ahead to the next unit even though I won’t have gotten a perfect score and had to look up a few questions. For the most part I think have a fairly strong grasp on integrals, plus I think that moving ahead will actually help  me better understand integrals when I inevitably have to use them in the math after this. Not to mention that I’ve been working through Integrals (3,070/3,200 M.P.) for 5 👏🏼 MONTHS 👏🏼….. So ya, my fingers are crossed that I get the next 4 questions correct and can finally (FIN-A-LLY) move on to the next unit, Differential Equations (0/1,100 M.P.).