Week 151 – June 18th to July 24th

I got smoked this week on KA. I didn’t get through the Integration Techniques unit test until Friday and spent Wednesday to Saturday trying to understand ONE question… And used 52 pages of notes trying to understand this question. 😳 I’m happy to say that I FINALLY figured out how to solve the question after spending around 7-8 hours working on it. I had a bit of a defeatist moment on Wednesday when I felt like I might have finally hit a wall and wasn’t going to be able to get past this point in calculus. The KA answer was insanely oversimplified and wasn’t any help so I tried using Symbolab but it also didn’t show me the steps to solve the question (Symbolab asks you to pay for a yearly subscription to see the steps of each equation). Finally (FINALLY!) I was able to grind my way through it and fully understand how/why it works. Since I didn’t actually learn anything new this week, the following are 4 questions from the test. (The really difficult one I just referred to is the last one.)

Question 1

I made a note after this question stating that I actually didn’t reach the proper solution to this question on my own before answering it. I realized that in order to solve it I needed to use integrations by parts but apparently, I messed it up. (To be honest, I completely forget working trough this question.) The third answer on KA was the closest one to the solution I reached on my own which is why I went with it and ended up getting lucky and got it correct.

Question 2

As I’m writing this, I just realized how insanely difficult this question was too. As far as I can tell and as you can see from my notes, before being able to come up with the solution to this question you already have to know what the derivative of tan(x) is, what the derivative of arctan(x) is, and what the derivative of 1/k * arctan((x + h)/k) is. If you know all of those things, the next step is to use the ‘completing-the-square’ method to simplify the denominator in the integrand. Lastly, you then plug the formulas into the integral and out pops the solution in about 4 steps. I find these questions super frustrating as I don’t have the trig identities or formulas memorized so I get stuck. I’m hoping that eventually I’ll have an intuitive understanding of the identities which will then make these types of questions simple to work through. 🙏🏼

Question 3

I was really happy with myself that I was able to solve this question in about 3 minutes. When I first looked at it I couldn’t remember the exact technique needed to work through it but had a vague recollection/intuition that I needed to break up cos3(x) into cos2(x)cos(x). I then somehow managed to remember that I could switch cos2(x) to (1 – sin2(x)) – which comes from the Pythagorean Theorem – and then multiply the two terms and use u-substitution on cos(x)sin2(x). Boom. 🧨

Question 4 (The mother of all questions)

First off, after 1.5 hours working on this question on Wednesday and getting nowhere, I was so agitated that when I tried to screen shot the question I clicked the wrong spot on KA, was taken away from the unit test, and then wasn’t able to screen shot it. 🤦🏻‍♂️ The question asked me to find the antiderivative of ∫1/(x2 + 4)3/2 * dx. The answer given to me on KA was about 7-8 steps which – as you’ll see from my notes below – is laughably over-simplified. Before you can begin to answer this question, there are a few formulas you need to know. One of them is what the derivative of tan(x) is which is a part of my notes from the second question above. Here’s the math to solve it:

  • d/dx[tan(x)] =
    • = d/dx[sin(x)/cos(x)]
    • = d/dx[(sin(x)(cos(x))-1]
      • a(x) = sin(x)
      • a’(x) = cos(x)
      • b(x) = x–2
      • b’(x) = –x–2
      • c(x) = cos(x)
      • c’(x) = –sin(x)
    • = d/dx[a(x)b(c(x))]
    • = a’(x)b(c(x)) + a(x)b’(c(x))c’(x)
    • = cos(x) * 1/cos(x) + sin(x)( –1/(cos(x))2)(–sin(x))
    • = cos(x)/cos(x) + sin2(x)/cos2(x)
    • = cos(x)/cos(x) * (cos(x)/cos(x))+ sin2(x)/cos2(x)
    • = cos2(x)/cos2(x)+ sin2(x)/cos2(x)
    • = (cos2(x) + sin2(x))/cos2(x)
    • = 1/cos2(x)
    • = sec2(x)

To solve ∫1/(x2 + 4)3/2 * dx, you also need to know that x can be replaced with 2tan(θ), and that (x2 + 4) can be replaced with 4sec2(θ). Here’s the proof for both:

  • sin2(θ) + cos2(θ) = 1
  • (sin2(θ) + cos2(θ))/cos2(θ) = 1/cos2(θ)
  • tan2(θ) + 1 = sec2(θ)
  • 4(tan2(θ) + 1) = 4(sec2(θ))
  • 4tan2(θ) + 4 = 4sec2(θ)
  • 22tan2(θ) + 4 = 4sec2(θ)
  • (2tan(θ))2 + 4 = 4sec2(θ)
    • Let 2tan(θ) = x
  • (x)2 + 4 = 4sec2(θ)

Once you state that x = 2tan(θ), you then need to set dx in terms of dθ:

  • d/dx[x] = d/dx[2tan(θ)]
    • 1 = 2 * d/dx[tan(θ)]
      • a(x) = tan(x)
      • a’(x) = sec2(x)
      • b(x) = θ
      • b’(x) = dθ/dx
    • 1 = 2 * d/dx[a(b(x))]
      • = 2 (a’(b(x)) * b’(x))
      • = 2 (sec2(θ) * dθdx)
    • dx(1) = dx(2sec2(θ) * dθdx)
    • dx = 2sec2(θ)dθ

Then, after getting all of that out of the way (which I guess you’re just supposed to know that you’re supposed to do? 🤔), you plug everything into the integral:

  • ∫1/(x2 + 4)3/2 * dx =
    • = ∫ (1/(x2 + 4))3/2 dx
      • Replace x with 2tan(θ) and dx with 2sec2(θ)dθ
    • = ∫ (1/((2(tan(θ))2 + 4)3/2 * 2sec2(θ)dθ
    • = ∫ (1/(4(tan2(θ) + 4)3/2 * 2sec2(θ)dθ
      • Replace 4(tan2(θ) + 4) with 4sec2(θ)
    • = ∫ (1)3/2/(4(sec2(θ))3/2 * 2sec2(θ)dθ
    • = ∫ 1/(22(sec2(θ))3/2 * 2sec2(θ)dθ
    • = ∫ 1/((2sec2(θ))2)3/2 * 2sec2(θ)dθ
    • = ∫ 1/(2sec(θ))3 * 2sec2(θ)dθ
    • = ∫ 1/(8sec3(θ)) * 2sec2(θ)dθ
    • = ∫ 2sec2(θ)/(8sec3(θ)) * dθ
    • = ∫ 1/(4sec(θ) *dθ
    • = ¼ ∫ 1/(1/1/cos(θ)) *dθ
    • = ¼ ∫ cos(θ)dθ
    • = ¼ sin(θ) + C
      • Replace θ with arctan(x/2)
    • = ¼ sin(arctan(x/2)) + C

When you reach the point that ∫1/(x2 + 4)3/2 * dx = ¼ sin(arctan(x/2)) + C, you’re not done yet! You need to simplify sin(arctan(x/2)). How? Simple! With this identity:

I’m not going to type out the math since you need to draw a triangle to understand sin(arcsin(x)), which I don’t think is possible on Word. Once you understand that ¼ sin(arctan(x/2)) = x/(x2 + 4)1/2 you can then finish off the integral:

  • ∫1/(x2 + 4)3/2 * dx =
    • = ¼ sin(arctan(x/2)) + C
    • = ¼ * x/(x2 + 4)1/2
    • = x/4(x2 + 4)1/2

Even though I got my ass kicked this week, I’m now feeling more confident than ever in my ability to work through trig identity problems. I spent hours working through the algebra from the last question and now have a MUCH stronger grasp on algebra, trig, and calculus having struggled through it. Unfortunately, considering how much practice I need with trig substitution, this means that I likely won’t get through Calc.2 by September. 😔 My plan is to rewatch all the 8 videos from the Trigonometric Substitution section of Integration Techniques (1,010/1,100 M.P.) which will probably take me almost a week to do alone. As disappointed as I am knowing that it’s likely going to take me longer than I’d hoped to get through Calc.2, if I can FINALLY get an intuitive feel for trig identities, it will be worth it. Plus I’ll feel like an absolute gangster and be incredibly relieved/proud of myself for getting through it. 💪🏼