For the first time in what feels like forever, I’m actually happy with the progress I made this week on KA. It took me three days but I finally got through the exercise in the section Trigonometric Substitution and was able to wrap my head around how this particular type of trig substitution works (with one frustrating exception which I’ll talk about below). After finishing that exercise, I then moved on to the following section, Integration by Parts, and watched all 6 videos in it which were pretty straightforward and easy to understand. I was happy to start feeling more comfortable with integration by parts since 2 weeks ago I watched a few of videos on it and was fairly confused by it. I still need a lot of practice using this method, however, but I made a lot of progress towards understanding it this week. 👍🏻 WOO!!

Working through the trig sub exercise this week, I really started to understand how to use a right-triangle to come up with Sine, Cosine, and Tangent functions (and their inverses) to use in place of expressions within an integral. Here are two questions I worked through from he exercise:

            Question 1

• ∫ 1/(x²√(4 + x²) dx = ∫ 1/(x²√(2² + x²) dx
  • (Note: On the first page of my notes, you can see that I drew a triangle on the left side of the photo. Knowing that a² + b² = c², the expression in the denominator of the integrand has the shape of c = √(a² + b²) so you’re able to label they hypotenuse as such, the adjacent side as 2 and the opposite side as x. The goal from there is to use trig functions to isolate √(2² + x²) and replace that expression in the integrand with a trig function. In this particular case, the replacement ended up being 2sec(θ) =√(2² + x²). Using this method, you also need to solve for x and dx which were x = 2tan(θ) and dx = 2sec²(θ)dθ.)
    • c = √(a² + b²)
    • 2sec(θ) =√(2² + x²)
    • x = 2tan(θ)
    • dx = 2sec²(θ)dθ
• ∫ 1/(x²√(2² + x²) dx = ∫ 1/(((2tan(θ))² * 2sec(θ)) * 2sec²(θ)dθ
  • = ∫ 1/((4 * tan²(θ) * 2sec(θ)) * 2sec²(θ) dθ
  • = ∫ ¼ * 1/(tan²(θ)) * sec(θ) dθ
  • = ¼ ∫ 1/(sin²(θ)/cos²(θ)) * (1/cos(θ)) dθ
  • = ¼ ∫ (cos²(θ)/sin²(θ)) * (1/cos(θ)) dθ
  • = ¼ ∫ (cos²(θ)/sin²(θ)) * (1/cos(θ)) dθ
  • = ¼ ∫ (cos²(θ)/sin²(θ)) * (1/cos(θ)) dθ
  • = ¼ ∫ cos(θ)/sin²(θ) dθ
  • = ¼ ∫ cos(θ) * sin^-2(θ) dθ
    • a(x) = x^-1/-1 = -x^-1
    • a’(x) = x^-2
    • b(x) = u = sin(θ)
    • b’(x) = du = cos(θ) * dθ
  • = ¼ ∫ -u^-2 dθ
  • = ¼ (–sin^-1(θ)) + C
  • = ¼ (-1/sin(θ)) + C
    • (Note: Looking back at the triangle, sin(θ) = x/√(2² + x²).)
  • = ¼ (-1/(x/(√(2² + x²)))) + C
  • = –¼ (√(2² + x²)/x) + C
  •  = –√(2² + x²)/4x + C

Question 2

This question highlights the issue I have with this type of trig substitution which is that I don’t know to tell which value should be assigned to the adjacent side or opposite side of the right-triangle. In the photo above, you can see that I drew 2 triangles where the values for the adjacent and opposite sides are swapped. Depending on which triangle you use, you’ll get different values for sin(θ), cos(θ), etc. In these questions, I’m not sure how to tell from the start which triangle you should go with to find the proper solution.

Here’s an example of a question where I drew the wrong triangle and came to the wrong conclusion as a result:

In the first photo, you can see on the left-hand side of the page that I drew a triangle where I labelled the adjacent side as √(x² – 4) which lead me to the conclusion that the integral ₂∫⁴ √(x² – 4)/x dx could be simplified to ­­ ₂∫⁴-2 cot²(θ) dθ. It turned out, however, that I was supposed to draw the triangle the other way around, as shown on the second page, which resulted in the integral equaling ₂∫⁴ 2 tan²(θ) dθ. As I said, I have no idea how to know which triangle I should go with from the start which is very frustrating. 😡

Regardless, I managed to get through the exercise on Thursday and began the section Integration by Parts on Friday. The first thing this section went over was that the method ‘integration by parts‘ is “the reverse power rule”, as Sal said. The formula is ∫ a’(x)b(x) = a(x)b(x) – ∫ a(x)b’(x) with the proof being:

• d/dx[a(x)b(x)] = a’(x)b(x) + a(x)b(x)
  • ∫ d/dx[a(x)b(x)] = ∫ a’(x)b(x) + ∫ a(x)b’(x)
  • ∫ d/dx[a(x)b(x)] = ∫ a’(x)b(x) + ∫ a(x)b’(x)
  • a(x)b(x) = ∫ a’(x)b(x) + ∫ a(x)b’(x)
  • a(x)b(x) – ∫ a(x)b’(x) = ∫ a’(x)b(x)
  • ∫ a’(x)b(x) = a(x)b(x) – ∫ a(x)b’(x)

Once I went through this proof a few times, this formula seemed very straightforward and easy to understand. That being said, I still found the questions I worked through in the other 5 videos a bit tricky to solve since I don’t have any experience using this method. I’m not going to go through the math behind each question, but here are 3 questions I worked through that used this method:

I’m looking forward to this coming week and am optimistic that I’ll be able to get a more intuitive feel for how integration by parts works. I have two articles and two exercises left to get through in this section so I’m hoping I’ll be able to get through all four by the middle of the week so I can finally get back to the exercise Integration Using Trigonometric Identities from the unit Integrals (2,530/3,200 M.P.). If all goes according to plan, I think I should be able to FINALLY get through Integrals in the next 2-3 weeks (🤞🏼). Getting through this unit has been a long time in the making so, when I finally do manage to get through it, it will definitely be a huge relief. 🥵