Week 125 – Jan. 17th to Jan. 23rd

(Disclaimer: This was a very frustrating week and so I swear a lot in this post. In other posts I’ve written, I’ll often write a first draft which might include swear words but I’ll edit them out before I post it. I’ve decided to leave them in this week because 1) it paints a more honest picture of how I’m feeling right now, and 2) I think that it’s important to emphasize that learning math does not always come easy at times and that it can be SUPER frustrating. 😡 That said, I do end on a more optimistic note.)

I’m so fucking frustrated with this Trigonometric identity section… I feel like I’ve been completely thrown to the wolves without being properly prepared for how to work through these questions. I did make some progress this week but, even still, I can barely wrap my head around the algebra when the answer is GIVEN to me, let alone figure out the answers on my own…. The most frustrating part is that when I look at the solutions, the answers skip a million steps and expect me to somehow know what’s going on… At the beginning of the week, I thought I was making progress in this section and thought I might be able to get through this exercise (after 5 fucking weeks of working on it…). I was so incredibly wrong about that. This is the question I just worked on:

First off, I had no intuitive sense that I should have used the integration by parts technique on this integral. I first tried splitting sec3(x) into 1/cos(x) * 1/cos2(x) and then stating that cos2(x) = (1 – sin2(x)) and then tried expanding the denominator. This was not even close to the right way to solve it. I then tried to think of sec3(x) as sec(x) * sec2(x) and whether I could somehow use u-substitution, but that also didn’t work. I finally just gave up and asked for the answer which was when I found out that I should have used integration by parts. Here are the first few steps of how to evaluate this integral:

  • ∫ sec3(x) dx = ∫ sec(x)sec2(x) dx
    • = ∫ a(x) * b’(x) dx
      • a(x) = sec(x)
      • a’(x) = tan(x)sec(x)
      • b(x) = tan(x)
      • b’(x) = sec2(x)
        • (Right off the bat, I don’t have the derivatives of sec(x) and tan(x) memorized so I was pretty much fucked right from the beginning.)
    • = ∫ a(x) * b’(x) dx = a(x)b(x) – ∫ a’(x)b(x) dx
      • = sec(x) * tan(x) – ∫ tan(x)sec(x) * tan(x) dx
      • = sec(x) * tan(x) – ∫ tan2(x)sec(x) dx
        • (From here you have to have it memorized that tan2(x) = sec2(x) – 1 and somehow know that that’s the identity you should use at this point…)
      • = sec(x) * tan(x) – ∫ (sec2(x) – 1) * sec(x) dx
      • = sec(x) * tan(x) – ∫ sec3(x) – sec(x) dx
      • = sec(x) * tan(x) – ∫ sec3(x) + ∫ sec(x) dx
        • (Here, you have to remember that the left side of the equation is ∫ sec3(x) so you can add that integral to both sides of the equation to eliminate it from the right side. Obvious, right?) 
        • (Oh, and also! Don’t forget that in the step above, the negative sign in front of the third integral, ∫ sec(x) dx, needs to change to a positive!)
      • 2 * ∫ sec3(x) dx = sec(x) * tan(x) + ∫sec(x) dx
        • (Here’s the best part! At this point you somehow need to know that the antiderivative of ∫sec(x) dx = ln|sec(x) + tan(x)|. How are you supposed to know? Obviously you’re just supposed to have it memorized even though it’s NEVER BEEN TAUGHT TO ME UP TO THIS POINT. Just in case you’re an idiot like me, here’s the process of how to solve it which, in case you were wondering, I found on Wikipedia.)

(To be clear, to find the antiderivative of ∫sec(x) dx I guess I was just supposed to assume that I needed to multiply sec(x) by ((sec(x) + tan(x))/((sec(x) + tan(x)). Then after that, I should have also realized that since the derivative of tan(x) is sec2(x) and the derivative of sec(x) is tan(x)sec(x) that I could simply set u to be equal to sec(x) + tan(x) and du to be equal to (sec2(x) + sec(x) + tan(x)) which simplifies the integral to 1/u. I should have realized that’s what I was supposed to do even though I’ve never seen a u-substitution where there are two terms being replaced by a single u in a denominator and two terms being replaced by du in the numerator. From there, obviously the antiderivative of 1/u is ln(|u|).)

  • (…)
    • (…)
      • (..)
        • (So… now that you understand that the clear and obvious antiderivative of ∫sec(x) dx is ln|sec(x) + tan(x)|, it’s easy to solve the rest of the problem.)
      • 2 * ∫ sec3(x) dx = sec(x) * tan(x) + ln(|sec(x) + tan(x)|)
      • ∫ sec3(x) dx = ½ *sec(x) * tan(x) + ½ * ln(|sec(x) + tan(x)|)

So simple, right?!?!?!

I’m honestly so fucking fed up with these questions and having to 1) have all the trig identities memorized, 2) somehow know which trig identity to use and when, 3) know when to use integration by parts, 4) always remember the derivatives and antiderivatives of cot(x), sec(x), tan(x), 5) know how/when to use u-substitution, AND ALL THE REST OF THE FUCKING RULES!!!!

FUCK THESE FUCKING TRIG FUCKING IDENTITY FUCKING QUESTIONS

😡😡😡😡😡😡😡😡

AAAAAAAAGGGGGGHHHHHHHGGHH!!!!!!!!!

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I honestly think I might cry……

😔

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I did manage to get some things done/figured out this week… 

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I finished off the integration by parts exercise on Tuesday as I was hoping I would. I was happy that I didn’t find it all that difficult. I also had a bit of success working through a couple of the trig sub questions. Here are 3 of them:

Question 1:

  • ∫ tan(x) dx = ∫ sin(x)/cos(x) dx
    • = ∫ sin(x) * (cos(x))-1 dx
    • = ∫ –1 * –1 * sin(x) * (cos(x))-1 dx
    • = –1 * ∫ * –1 * sin(x) * (cos(x))-1 dx
    • = –1 * ∫ –sin(x) * (cos(x))-1 dx
    • = –1 * ∫ b’(x) * a’(b(x)) dx
      • a(x) = ln(x)
      • a’(x) = x-1 = 1/x
      • b(x) = cos(x)
      • b’(x) = –sin(x)
    • = –1 * (a(b(x)) + C
    • = –1 * (ln(|cos(x)|) + C
    • = –1 * ln(|cos(x)|) + C
    • = ln(|cos(x)-1| + C
      • (Using Log rules, the –1 in front of ln(|cos(x)|) gets turned into an exponent on cos(x).)
    • = ln(|1/cos(x)| + C
    • = ln(|sec(x)| + C

Question 2:

  • ∫ sin4(x) dx = ∫ (cos2(x))2 dx
    • = ∫ ((1 – cos(2x))/2)2 dx
      • (This is the double angle identity where sin2(x) = ((1 – cos(2x))/2. I could be wrong, but I believe that when there’s a sin or cos function being raised to an even power, using the double angle identity is usually what you want to do.)
    • = ∫ (½ – cos(2x)/2)2 dx
    • = ∫ (½ – cos(2x)/2) * (½ – cos(2x)/2) dx
    • = ∫ ¼ – (2cos(2x))/4 + cos2(2x)/4 dx
    • = ∫ ¼ (1 – 2cos(2x) + cos2(2x)) dx
    • = ¼ * ∫ 1 – 2cos(2x) + cos2(2x) dx
      • (Here you have to use a double angle identity again but this time you use the cosine double angle identity where cos2(x) = ((1 + cos(2))/2) BUT you have to remember that the angle switches to 4x since the original angle is 2x, i.e. 2(2x) = 4x.)
    • = ¼ * ∫ 1 – 2cos(2x) + (½ + cos(2(2x))/2 dx
    • = ¼ * ∫ 1 – 2cos(2x) + ½ + cos(4x)/2 dx
    • = ¼ * ∫ 3/2x0 – 2cos(2x) + cos(4x)/2 dx
      • (The ½ gets added to the 1 turning it into 3/2. I also added a x0 onto it for the next step.)
    • = ¼ * ∫ 3/2x0 dx – ¼ * ∫ 2cos(2x) dx + ¼ * ∫ cos(4x)/2 dx
    • = ¼ * ∫ 3/2x0 dx – ¼ * ∫ 2cos(2x) dx + ¼ * ∫ 4/4 * cos(4x)/2 dx
    • = ¼ * ∫ 3/2x0 dx – ¼ * ∫ 2cos(2x) dx + ¼ * ∫ 1/8 * 4 * cos(4x) dx
    • = ¼ * ∫ 3/2x0 dx – ¼ * ∫ b’(x) * a’(b(x)) dx + ¼ * ∫ 1/8 * c’(x) * a’(c(x)) dx
      • a(x) = sin(x)
      • a’(x) = cos(x)
      • b(x) = 2x
      • b’(x) = 2
      • c(x) = 4x
      • c’(x) = 4
    • = ¼ * 3/2x – ¼ * a(b(x)) + ¼ * 1/8 a(c(x))  + C
    • = ¼ * (3/2x – a(b(x)) + 1/8 * a(c(x))) + C
    • = ¼ * (3/2x – sin(2x)) + 1/8 * sin(4x)) + C

(What pisses me off beyond belief is that the answer shown in the second screen shot above simplifies everything I just wrote out to 4 steps…. FOUR FUCKING STEPS. It’s honestly SO infuriating…)

Question 3:

  • 3π/4∫3π/2 sin2(x/3) dx = 3π/4∫3π/2 ((1 – cos(2x/3))/2) dx
    • = 3π/4∫3π/2 (½ – cos(2x/3)/2) dx
    • = 3π/4∫3π/2 ½x0 – ½(cos(2x/3) dx
    • = 3π/4∫3π/2 ½x0 – ½(cos(2x/3) dx
    • = 3π/4∫3π/2 ½x0 – ½(cos(2x/3) * 3/2 * 2/3 dx
    • = 3π/4∫3π/2 ½x0 – ½ * 3/2 *(cos(2x/3) * 2/3 dx
    • = 3π/4∫3π/2 ½x0 – 3/4 * (cos(2x/3) * 2/3 dx
    • = 3π/4∫3π/2 ½x0 – 3/4 * a’(b(x)) * b’(x) dx
      • a(x) = sin(x)
      • a’(x) = cos(x)
      • b(x) = 2x/3
      • b’(x) = 2/3
    • = ½x – 3/4 * a(b(x)) 3π/4|3π/2
    • = ½x – 3/4 * sin(2x/3) 3π/4|3π/2
      • A(x) b|c = A(c) – A(b)
        • A(x) = ½x – 3/4 * sin(2x/3)
        • c = 3π/2
        • b = 3π/4
    • = (½(3π/2) – 3/4 * sin(2(3π/2)/3)) – (½(3π/4) – 3/4 * sin(2(3π/4)/3))
    • = (½(3π/2) – 3/4 * sin(π)) – (½(3π/4) – 3/4 * sin(π/2))
    • = (3π/4) – 3/4 * sin(π)) – (3π/8 – 3/4 * sin(π/2))
    • = (3π/4) – 3/4 * (0)) – (3π/8 – 3/4 * (1))
    • = (3π/4) – (3π/8 – 3/4)
    • = 3π/4 – 3π/8 + ¾
    • = 6π/8 – 3π/8 + 6/8
    • = 3π/8 + 6/8
    • = (3π + 6)/8

As pissed off as I am with how things are going right now, in a weird way I’m even more fired up to get through this. I feel like I’m getting my ass kicked by some god of math, but there’s no fucking way I’ve come this far to give up now. I’m 100% going to get through this stupid fucking section on trig substitution. I’m hoping I can get through it this coming week and finally (FIN-A-LLY) get through Integrals (2,480/3,200 M.P.). For the love of God, please (PLEASE!!!!!!!!!) let that be the case. 🙏🏼 🙏🏼 🙏🏼 🙏🏼 🙏🏼 🙏🏼 🙏🏼