I’m starting to feel more comfortable with trigonometric identities but at the same time still feel like I have a long way to go before I fully understand them. As I mentioned at the end of my last post, this week I took a break from the unit Integrals and worked through a section in the unit Integration Techniques in the subject Calculus 2. The section I worked on is called Trigonometric Substitution. I managed to get through all 8 videos in this section and technically completed the exercise by getting 4/4 question correct, however I guessed on one of the questions and happened to get it correct so I’m planning to redo the exercise next week so that I get more comfortable with these types of questions. I only learned about a handful of new concepts this week which I’ll explain by going through 4 example questions.

The main thing I learned this week was how to use a triangle to replace terms in an integral with trig identities – shocking given that the title of this section is “Trigonometric Substitution” /s. The very first video I watched this week went through this process step by step. Here are the notes I took from the first video with an explanation of each step below. Below that is a question I worked through in the exercise that used this concept.

            Question 1 (from video):

• Step 1 – Write out integral so that the denominator takes the form of √(c² – a²), a.k.a. the Pythagorean Theorem rewritten to solve for side length b.
  • ∫ 1/(√4 – x²) dx = ∫ 1/(√2² – x²) dx
• Step 2 & 3 – Draw diagram of right-triangle and label each side with values from integral.
  • (See diagram above.)
• Step 4 – Determine opposite and adjacent side lengths with appropriate sin(θ) and cos(θ) values using SOH and CAH.
  • sin(θ) = x/2
    • x = 2*sin(θ)
  • cos(θ) = adj/hyp
    • = √(2² – x²)/2
    • √(2² – x²) = 2 * cos(θ)
• Step 5 – Replace opposite and adjacent length values on diagram with trig values and determine dx.
  • (See diagram above where I labelled the new triangle side-length values.)
  • x = 2*sin(θ)
  • dx = 2 * cos(θ) * dθ
• Step 6 – Replace denominator and dx in integral with trig substitutes, cancel terms and evaluate integral.
  • ∫ 1/(√2² – x²) dx = ∫ 1/(2 * cos(θ)) * 2 * cos(θ) * dθ
    • = ∫ 1/(2 * cos(θ)) * 2 * cos(θ) * dθ
    • = ∫ 1/2 * 2 * dθ
    • = ∫ dθ
    • = θ + C
• Step 7 – Determine value of θ in terms of x and replace θ in solution.
  • ∫ 1/√(2² – x²) dx = θ + C
    • sin(θ) = x/2
      • (This was from Step 4.)
    • θ = sin^-1(x/2)
  • ∫ 1/√(2² – x²) dx = sin^-1(x/2) + C

Question 2 (from exercise):

(Note: This was the question I guessed on by picking a random answer. I didn’t actually know the correct answer when I checked it.)

• Q. Evaluate ∫ √(25x² – 4)/x * dx
  • (Note: The key here is to recognize that the numerator can take the shape of √(c² – a²) where c² = 25x² and b² = x². From there, as you can see in the bottom left corner on the first page of my notes, you can then draw a triangle and relabel each side using the terms in the numerator and come up with trig values to use in the integrand.)
  • (Note #2: Since c² = 25x² then √c² = √(25x²) -> c = 5x which you can see that I labelled on the triangle.)
  • 5x/2 = hyp/adj = sec(θ)
    • x = (2/5)*sec(θ)
  • dx = (2/5)*tan(θ)*sec(θ)dθ
• ∫ √(25x² – 4)/x * dx = ∫ √((25((2/5)*sec(θ))² – 4))/(2/5)*sec(θ) * (2/5)*tan(θ)*sec(θ)*dθ
  • = ∫ √((25(4/25)*sec²(θ) – 4))/(2/5)*sec(θ) * (2/5)*tan(θ)*sec(θ)*dθ
  • = ∫ √((4*sec²(θ) – 4))/(2/5)*sec(θ) * (2/5)*tan(θ)*sec(θ)*dθ
  • = ∫ √((4*sec²(θ) – 4))/(2/5)*sec(θ) * (2/5)*tan(θ)*sec(θ)*dθ
  • = ∫ √((4*sec²(θ) – 4)) * tan(θ) *dθ
  • = ∫ √(4(sec²(θ) – 1)) * tan(θ) *dθ
    • (Note: In the step just above I factored out a 4 from the two terms in the numerator leaving me with sec²(θ) – 1 which equals tan²(θ) because:
      • cos²(θ) + sin²(θ) = 1
        • (cos²(θ)/cos²(θ)) + (sin²(θ)/cos²(θ)) = 1/cos²(θ)
        • 1 + tan²(θ) = sec²(θ)
        • tan²(θ) = sec²(θ) – 1
    • Boom.)
  • = ∫ √(4(tan²(θ))) * tan(θ) * dθ
  • = ∫ 2 * tan(θ) * tan(θ) * dθ
  • = ∫ 2 * tan²(θ) *dθ
  • = ∫ 2 * (sec²(θ) – 1) * dθ
  • = ∫ 2sec²(θ) – 2θ⁰ * dθ
    • (Note: In the above step I expanded 2 * (sec²(θ) – 1) so that it became 2sec²(θ) – 2 and then added a θ⁰ to the second term so that I could then use the reverse power rule in the following step.)
  • = 2tan(θ) – 2θ + C
    • (Note: Looking at the triangle, the opposite side equals √(25x² – 4) and the adjacent side equals 2 so therefore 2tan(θ) = 2(opp/adj) = 2(√(25x² – 4))/2 = √(25x² – 4).)
  • = √(25x² – 4) – 2*sec^-1(5x/2) + C
    • (Note: In my notes I wrote the answer as √(25x² – 4) – 2*cos^-1(2/5x) + C where I used arccos instead of arcsec but it’s the same thing.)

One of the trig identities/substitutions I learned about this week had to with identifying an expression inside an integrand as a² – x² so that you can substitute that expression with a²cos²(θ). The proof for this substitution is:

• a² – x²
  • *Substitute x with asin(θ)*
    • (Note: This part still confuses me. It seems weird to me that you can just all of a sudden decide to let x = asin(θ).)
• a² – x² = a² – (asin(θ))²
  • = a² – a²sin²(θ)
  • = a²(1 – sin²(θ))
  • = a² * cos²(θ)

Here’s a question from my notes where I used this identity/substitution:

• ∫ x³ * √(9 – x²) dx = ∫ x³ * √(3² – x²) dx
  • (Note: Underneath the radical where it states 3² – x² is where you can use the substitution a² – x² = a² * cos²(θ).)
• ∫ x³ * √(3² – x²) dx = ∫ x³ * √(3²cos²(θ)) dx
  • = ∫ x³ * 3 * cos(θ) dx
    • (Note: Remembering that in the proof of a² – x² = a² * cos²(θ) it was stated that x = asin(θ), in the next step you substitute x in the term x³ with asin(θ) where a = 3.)
    • (Note #2: At this point, you should also replace dx with the derivative of x in terms of θ [not sure if I’m saying the right] which is 3*cos(θ)*dθ.)
  • = ∫ (3*sin(θ))³ * 3 * cos(θ) * 3*cos(θ)*dθ
  • = ∫ 27*sin³(θ) * 3 * cos(θ) * 3*cos(θ)*dθ
  • = ∫ 243 * sin³(θ) cos²(θ) dθ
  • = 243 ∫ sin²(θ) * cos²(θ) * sin(θ) * dθ
  • = 243 ∫ (1 – cos²(θ)) * cos²(θ) * sin(θ) * dθ
    • (Note: Here you need to expand (1 – cos²(θ)) * cos²(θ).)
  • = 243 ∫ cos²(θ) – cos⁴(θ) * sin(θ) * dθ
    • cos(θ) = u
    • –sin(θ)* dθ = du
      • (Note: the derivative of cos(θ) is –sin(θ) so in the step below you multiply 243 and sin(θ) both by –1 which doesn’t change the value of the expression and allows you to state that –sin(θ)* dθ = du.)
  • = –243 ∫ cos²(θ) – cos⁴(θ) * –sin(θ) * dθ
  • = –243 ∫ u² – u⁴ * du
    • *Reverse Power Rule*
  • = –243 * (u³/3 – u⁵/5) + C
  • = –243 * (cos³(θ)/3 – cos⁵(θ)/5) + C
    • x = asin(θ) = 3sin(θ)
      • sin(θ) = x/3
    • cos²(θ) + sin²(θ) = 1
      • cos(θ) = √(1² – sin²(θ))
        • √(1 – (sin(θ))²)
  • = –243 * ((cos(θ))³/3 – (cos(θ))⁵/5) + C
  • = –243 * ((√(1 – (sin(θ))²))³/3 – (√(1 – (sin(θ))²))⁵/5) + C
  • = –243 * ((√(1 – (sin(θ))²))³/3 – (√(1 – (sin(θ))²))⁵/5) + C
  • = –243 * ((√(1 – (x/3)²))³/3 – (√(1 – (x/3)²))⁵/5) + C
  • = –243 * ((√(1 – (x²/9)))³/3 – (√(1 – (x²/9)))⁵/5) + C
  • = –243 * ((1 – (x²/9)^3/2)/3 – (1 – (x²/9)^5/2)/5) + C
    • (Note: Since there’s a (–) sign in front of the constant, –243, the final step here is to multiply the constant and everything inside the parenthesis by –1. This in effect flips the order of the terms inside the parenthesis so that the term with the higher power, (1 – (x²/9)^5/2)/5, is subtracting the term with the lower power, (1 – (x²/9)^3/2)/3.)
  • = 243 * ((1 – (x²/9)^5/2)/5 – (1 – (x²/9)^3/2)/3) + C

Finally, there was one more question I worked through in the final video of this section that I thought was worth adding to this post. The video was titled Long Trig Sub Problem which is exactly what this question was. Below are the notes I took while answering it. Considering how long my notes are for this question and given how long it’s already taken me to write out the math on the previous 2 questions, I’m not going to go through the math here. Hopefully you’re able to decipher my chicken-scratch writing and follow along with my notes!

As you can tell, these trig sub questions are very involved and it seems like you have to have a pretty good grasp on trig identities in order to intuitively know when to substitute parts of the equations and which identity to use at what time. As I said at the beginning, I do think I’m getting better at it but still feel like I have a long way to go. 😔 One positive is that the Algebra seems to be less overwhelming than it was a few weeks ago. 👍🏻

As I also mentioned earlier, my plan for this coming week is to start by redoing the exercise from this section which I’m hoping will only take me a day or two to get through. After that, I’ll then get started in the following section Integration by Parts. It’d be nice to get through that section by the end of next week so I can get back to Integrals (2,530/3,200 M.P.) the week after, but I don’t know if that’s going to happen. My progress seems SUPER slow these days but, nonetheless, I’m still progressing which is what counts!