I only made it through 2 articles and 1 exercise this week from the unit Integration by Parts, but, just like last week, I’m actually fairly happy with the progress I made and what I was able to learn. I got a much stronger feel for how to use the integration by parts technique and came up with some good strategies to help me work through these types of questions. I also got a better feel for how to determine which term inside the integrand should be labelled as the derivative and which term should be considered the antiderivative (although I’m not sure if that’s actually the correct way to use those terms). The majority of the questions I worked through also were made up of composite functions inside the integrand so, on top of using integration by parts technique, I also got to practice using the reverse chain rule. All in all, I feel much less intimidated by these types of integrals so, as I said, I’m happy with how this week went.

Being that I didn’t actually learn anything new this week and only practiced working through questions, I don’t have anything new to talk about so I’m just going to go over 4 questions I worked on this week.

            Question 1

• ∫ x² * sin(πx) dx = ∫ a(x) * b(c(x)) dx
  • (Note: The key here, and with all the questions I worked through this week is to figure out which function should be considered the derivative and which one should be considered the antiderivative. In this instance, the best thing to do it to set x² as a(x) and sin(πx) as b’(x) and then adding a π/π, taking a 1/πout of the integrand, and then setting π to be equal to c’(x). From there you can use integration by parts.)
  • a(x) = x²
  • a’(x) = 2x
  • b(x) = –cos(x) 
  • b’(x) = sin(x)
  • c(x) = πx
  • c’(x) = π
• 1/π * ∫ x² * sin(πx) π dx = 1/π * ∫ a(x) * b’(c(x)) c’(x) dx
  • = 1/π * (a(x) * b(c(x) – ∫ a’(x) * b(c(x)) dx)
  • = 1/π * (x² * –cos(πx) – ∫ 2x * –cos(πx) dx)
  • = 1/π * (x² * –cos(πx) + 2 * (∫ x * cos(πx) dx))
    • (Note: At this point, you simply use integration by parts again and set x to be a(x) which makes it’s derivative in the following step equal to 1 and simplifies the integral.)
    • a(x) = x
    • a’(x) = 1
    • b(x) = sin(x)
    • b’(x) = cos(x)
    • c(x) = πx
    • c’(x) = π
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * ∫ x * cos(πx) * π dx))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * ∫ a(x) * b’(c(x) * c’(x) dx))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * (a(x) * b(c(x)) – ∫ a’(x) * b(c(x) * dx)))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * (x * sin(πx) – ∫ (1) * sin(πx) * dx)))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * (x * sin(πx) – ∫ sin(πx) * π/π dx)))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * (x * sin(πx) – 1/π * (∫ sin(πx) * π dx))))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * (x * sin(πx) – 1/π * (∫ a’(b(x) * b’(x) dx))))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * (x * sin(πx) – 1/π * (a(b(x))))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * (x * sin(πx) – 1/π * (–cos(πx))))
  • = 1/π * (x² * –cos(πx) + 2 * (1/π * (x * sin(πx) + (cos(πx)/π)))
  • = 1/π * (x² * –cos(πx) + 2 * ((x * sin(πx)/π) + (cos(πx)/π²))
  • = 1/π * (x² * –cos(πx) + ((2x * sin(πx)/π) + (2 * cos(πx)/π²))
  • = (x² * –cos(πx)/π) + (2x * sin(πx)/π²) + (2 * cos(πx)/π³)

Question 2

• ₀∫^π x * sin(2x) dx = ₀∫^π a(x) * b(c(x)) dx
  • a(x) = x
  • b(x) = sin(x)
  • c(x) = 2x
• ₀∫^π x * sin(2x) dx = ₀∫^π x * sin(2x) * 2/2 dx
  • = ½ * ₀∫^π x * sin(2x) * 2 dx
  • = ½ * ₀∫^π a(x) * b’(c(x)) * c’(x) dx
    • a(x) = x
    • a’(x) = 1
    • b(x) = –cos(x)
    • b’(x) = sin(x)
    • c(x) = 2x
    • c’(x) = 2
  • = ½ * (a(x) * b(c(x)) – ₀∫^π a’(x) * b(c(x)) dx)
  • = ½ * (x * –cos(2x) – ₀∫^π (1) * –cos(2x) dx)
  • = ½ * (x * –cos(2x) – ₀∫^π –cos(2x) * 2/2 dx)
  • = ½ * (x * –cos(2x) – ½ * ₀∫^π –cos(2x) * 2 dx)
  • = ½ * (x * –cos(2x) – ½ * ₀∫^π e’(f(x)) * f’(x) dx)
    • e(x) = –sin(x)
    • e’(x) = –cos(x)
    • f(x) = 2x
    • f’(x) = 2
  • = ½ * (x * –cos(2x) – ½ * (e(f(x))) ₀|^π
  • = ½ * (x * –cos(2x) – ½ * (–sin(2x)) ₀|^π
  • = ½ * (x * –cos(2x) – (–sin(2x)/2) ₀|^π
  • = ½ * (x * –cos(2x) + sin(2x)/2) ₀|^π
  • = ((x * –cos(2x)/2) + sin(2x)/4) ₀|^π
    • (Note: At this point you have to input both π  and 0 into the antiderivative separately and then subtract the value of the antiderivative when 0 is inputted from the value of the antiderivative when π is inputted.)
  • = ((π * –cos(2π)/2) + sin(2π)/4) – ((0 * –cos(2(0))/2) + sin(2(0))/4)
  • = ((π * –cos(2π)/2) + sin(2π)/4) – ((0 * –cos(2(0))/2) + sin(2(0))/4)
  • = (π * –cos(2π)/2)
  • = (π * –(1)/2)
  • = –π/2

Question 3

• ∫ x * e^-x dx = ∫ a(x) * b(c(x)) dx
  • a(x) = x
  • b(x) = e^x
  • c(x) = –x
• ∫ x * e^-x dx = ∫ x * e^-x * (–1/–1) dx
  • = –1 * ∫ x * e^-x * –1 dx
  • = –1 * ∫ a(x) * b’(c(x)) * c’(x) dx
    • a(x) = x
    • a’(x) = 1
    • b(x) = e^x
    • b’(x) = e^x
    • c(x) = –x
    • c’(x) = –1
  • = –1 * (a(x) * b(c(x)) – ∫ a’(x) * b(c(x)) dx)
  • = –1 * (x * e^-x – ∫ (1) * e^-x dx)
  • = –1 * (x * e^-x – ∫ e^-x * (–1/–1) dx)
  • = –1 * (x * e^-x – (–1) * ∫ e^-x * –1 dx)
  • = –1 * (x * e^-x + ∫ e’(f(x)) * f’(x) dx)
    • e(x) = e^x (😂)
    • e’(x) = e^x
    • f(x) = –x
    • f’(x) = –1
  • = –1 * (x * e^-x + e(f(x))
  • = –1 * (x * e^-x + e^-x)
  • = –e^-x * (x + 1)

Question 4

• ∫ ln(x) dx = ∫ ln(x) * x⁰ dx
  • = ∫ a(x) * b’(x) dx
    • a(x) = ln(x)
    • a’(x) = 1/x
    • b(x) = x
    • b’(x) = x⁰
  • = a(x) * b(x) – ∫ a’(x) * b(x) dx
  • = ln(x) * x – ∫ 1/x * x dx
  • = ln(x) * x – ∫ x/x dx
  • = ln(x) * x – ∫ 1 dx
  • = ln(x) * x – ∫ x⁰ dx
  • = (ln(x) * x – x) + C

I finished this week having gotten 2/4 questions correct on the final exercise of this unit so I’m hoping I’ll get through the remaining 2 questions on Tuesday and then be able to spend the rest of the week working through Integration Using Trigonometric Identities. I’m not all that confident that what I learned in the past two units is going to make it much easier for me to get through Integration Using Trigonometric Identities since, from what I can remember, that unit didn’t involve too much integration by parts. That said, I still think I’ll find it easier this time around having gotten quite a bit more practice working with integrals over the past few weeks. Hopefully!