After ~ 6 weeks, this week I FINALLY made it through that trig substitution section. 🥵 In fact, I actually got 4/4 on the exercise 3 times but wanted to make sure I knew what I was doing which is why I did it multiple times. Getting through this section has been the hardest thing I’ve thing I’ve done in KA over the past ~2.5 years. I now have a MUCH better understanding of integrals and how to integrate them, although I also know that I still have a long way to go before I understand them completely. Even though it took me such a long time to work through this section, I’m happy with everything I learned throughout the process. At the end of the week, I finally got started on the integral unit test and made it to the 5^th question which I ended up getting wrong so I stopped there. Regardless, I’m still pumped (and relieved) to finally be moving on.

Last week I had a hissy fit because I had never been shown what the antiderivative of ∫ sec(x) dx yet I was given a question where I was expected to know it somehow. This week, one of the first questions I worked on was specifically how to find the antiderivative of ∫ sec(x) dx… Here’s the question from KA and their solution:

Not long after I worked through that question, I was then given the same question I was given last week where I needed to know the antiderivative of ∫ sec(x) dx. Here’s the question and work I did to solve it below:

Since I already wrote out the math for this question in my last post, I’m not going to do it again here. The following are 2 other questions I worked through in the trig sub exercise this week. I forgot to take screen shots of both questions from KA so the screen shots are from the Symbolab calculator that I sometimes use:

Question 1:

• ∫ sin⁵(x) dx = ∫ sin(x) * sin²(x) * sin²(x) dx
  • = ∫ sin(x) * (1 – cos²(x)) * (1 – cos²(x)) dx
  • = ∫ sin(x) * (1 – 2cos²(x) + cos⁴(x)) dx
  • = ∫ sin(x) – 2sin(x)cos²(x) + sin(x)cos⁴(x) dx
  • = ∫ sin(x) + 2(–sin(x))(cos(x))² – (–sin(x))(cos(x))⁴ dx
    • (In the step above, I multiplied the middle and last term by (–1/–1) in order to turn sin(x) into –sin(x) in both terms.)
    • a(x) = x³/3
    • a’(x) = x²
    • b(x) = u = cos(x)
    • b’(x) = du = –sin(x)
    • c(x) = x⁵/5
    • c’(x) = x⁴
  • = ∫ sin(x) + ∫ 2 * b’(x) * a’(b(x)) – b’(x) * c’(b(x)) dx
  • = ∫ sin(x) dx + ∫ 2(u)² – (u)⁴ du
  • = –cos(x) + 2u³/3 – u⁵/5 + C
  • = –cos(x) + 2cos³(x)/3 – cos⁵(x)/5 + C

Question 2:

• ₀∫^π/2 3cos³(x) dx = 3 * ₀∫^π/2 cos²(x) * cos(x) dx
  • = 3 * ₀∫^π/2 (1 – sin²(x)) * cos(x) dx
  • = 3 * ₀∫^π/2 cos(x) – cos(x)sin²(x) dx
  • = 3 * ₀∫^π/2 cos(x) – cos(x)(sin(x))² dx
  • = 3 * ₀∫^π/2 cos(x) – 3 * ₀∫¹ u² du
    • (This is a step that I’m still trying to wrap my head around, i.e. using u-substitution and then changing the upper and lower bounds to the appropriate values.)
  • = 3 * sin(x) ₀|^π/2 – 3 * u³/3 ₀|¹
  • = 3 * sin(x) – 3 * sin³(x)/3 ₀|^π/2
  • = 3 * sin(x) – sin³(x) ₀|^π/2
    • A(x) _b|^c = A(c) – A(b)
      • A(x) = 3 * sin(x) – sin³(x)
      • c = π/2
      • b = 0
  • = (3 * sin(c) – sin³(c)) – (3 * sin(b)/2 – sin³(b))
  • = (3 * sin(π/2) – sin³(π/2)) – (3 * sin(0)/2 – sin³(0))
  • = (3 * sin(π/2) – sin³(π/2)) – (3 * sin(0)/2 – sin³(0))
  • = 3 * 1 – 1³
  • = 3 – 1
  • = 2

After I finally made it through the exercise, I started the unit test on Saturday. As I mentioned, I only made it through 5 questions and ended up getting the fifth question wrong. 😔 I was annoyed that I got a question wrong and extra annoyed that the question I got wrong had to do with Riemann Sums which, from what I remember, seemed like a pretty straightforward part of integrals. One thing I realized that made me feel a bit better, however, is that I first began this unit back in MID-OCTOBER… (😐) So I have sympathy for myself for not remembering what learned more than 3 months ago. In any case, here’s the question I got wrong:

Reading through the answer, I’m a bit confused and don’t really remember why the formula they give works. (FYI, the actual answer was (B).) Even though it’s frustrating that I don’t remember how Riemann sums work, I’m trying not to be too bummed about it and am reminding myself that it won’t take me too long to review them and remember how they work.

My goal for this coming week is to get through the unit test at least two times. There are 32 questions in the unit test so each attempt will likely take me at least two days. I won’t too upset if I don’t end up passing the test this week, especially considering how long it’s been since I first started this unit. My goal is simply to get refamiliarized with the material so that I can hopefully pass the test the following week. If that’s the case, it will end up being ~4 months of working through this unit, Integrals (2,560/3,200 M.P.), which will be a record for the most amount of time I’ve spent on a single unit. I remember when I first started KA there were weeks when I’d get through 3 units in a single week…