I finally made it through the Integration Techniques unit test this week! 🥳 I managed to rewatch all the videos in the Trigonometric Substitution section, then got through the final exercise in the unit that I needed to redo – Integration using ‘completing-the-square’ – and then finished off the unit test this morning. This week I got a MUCH better understanding of how and, more importantly, WHY trig-sub works which is a big relief. I still need more practice but I’m much more confident with trig-sub now. I also got a stronger grasp on u-sub which I was also very happy about. I still need a lot of practice with integration to quickly and intuitively recognize which techniques to use for each question in general, but I definitely made a lot of progress this week which is great.

It took me Monday, Tuesday, and Wednesday to get through the 8 trig-sub videos that I wanted to rewatch. After that I began the exercise Integration using ‘completing-the-square’ on Thursday. What I found annoying was that each question from the exercise used trig-sub yet the trig-sub section came AFTER the exercise… The order was so stupid! In any case, it took me 2 days to get through the exercise but it seemed much easier to understand this time around. Here are 2 questions from the exercise:

Question 1

These questions really helped me think through how to use a right triangle and use what I now think of as ‘theta-substitution’ for these types of questions. These questions also helped me understand and recognize quickly that if you see something like √(x² – 16) or √(x² + 16) you should think of it as potentially being b = √(c² – a²) or c = √(b² + a²). Here’s an explanation of the steps:

• Step 1
  • Use the complete-the-square method to turn –(x² –10x – 11) into 6² – (x – 5)², a.k.a. c² – a².
• Step 2
  • Input new expression back into denominator.
• Step 3 and 3.1
  • Draw a triangle and label each side accordingly based off the new expression.
• Step 3.2 and 3.3
  • Using the triangle, come up with values for x, θ, dx, and cos(θ).
• Step 4
  • Use theta-sub on expressions in integrand and simplify.
• Step 5
  • Like u-sub, replace θ with arcsin((x – 5)/6).

Question 2

This question highlights how you more-or-less need to have the derivative of arctan(x) memorized (d/dx[arctan(x)] = 1/(1 + x²)) so that you can quickly recognize it once you’ve used the completing-the-square method to get the integrand into the form of 1/(1 + x²). You can then use u-sub knowing that 1/(1 + x²) can be a’(x) and therefore arctan(x) = a(x). 

I started the Integration Techniques unit test on Friday and made it up to the 5^th question before having any trouble:

Question 3

The first way I sort of cheated was by looking at a list of derivatives and antiderivatives that I have saved to my desktop and saw that the antiderivative of tan(x) is ln|cos(x)|. I wasn’t sure if there was a simple antiderivative for tan(x) which is why I quickly checked but, since I ended up doing the math without any issues anyways, I could have figured it out on my own if I had taken the time. The second way I cheated was by looking up this equation:

I had ∫ tan(x)sec²(x) dx and I knew that sec²(x) is the derivative of tan(x) – I always think of it as ∫ b(x) * b’(x) dx – but didn’t know what to do from there. I looked up what the antiderivative of ∫sin(x)cos(x) dx was to get an idea of how to solve it. I’m glad I did because afterwards u-sub seemed more clear to me. I guess since you have ∫ b(x) * b’(x) dx you can just think of it as ∫ u¹ du which you can then integrate using the reverse power rule, u¹⁺¹/1+1 = u²/2 and then sub back in tan(x) = b(x) = u. This helps me to understand that sec²(x)*dx = du. This part was always really confusing to me so I’m glad I have a better idea of what’s going on now.

The next question I worked on, question 6 on the test, wasn’t hard but I needed to use partial fractions and I somewhat forgot how to do them. I eventually figured it out and was happy that I was able to get it on the first try:

Question 4

Question 7 on the test was the one question I legitimately cheated on and looked up before answering. I spent ~2 hours trying to work through it before looking up the solution. Here’s the question:

What I didn’t understand as I was trying to work through this question is you can imply that within ∫ sin^-1(x) dx there’s an x⁰. Knowing that, you can then use integration by parts followed by u-substitution to solve. Once I realized that I could add an x⁰, the next key was to recognize which term in the integrand was a derivative and which one was a function:

• ∫ sin^-1(x) * x⁰ dx
  • a(x) = sin^-1(x)
  • a’(x) = 1/(1 – x²)^1/2
  • b(x) = x
  • b’(x) = x⁰ = 1
• ∫ sin^-1(x) * x⁰ dx =
  • = ∫ a(x) * b’(x) dx
    • (Use integration by parts)
  • = a(x)b(x) – ∫ a’(x) * b(x) dx

As I said, from there you need to use u-sub which I didn’t find too difficult. I forgot to mention that, again, you essentially need to know or at least be able to figure out what the derivative of arcsin(x) is before you can answer this question.

I stared at the 8th question on the test for about 5 minutes before I made an attempt at answering it. Here it is:

Question 6

I thought this question was way more complicated than it actually was which is what stunned me at the beginning. Part of the reason why I froze was because I haven’t used the cosecant function in a long time and, although I remembered that cosecant is just the reciprocal function of sine, I assumed this question was harder than a regular sine, cosine, or tangent question. I started by writing out that csc = hyp/opp which led me to drawing a triangle and then started labelling the sides based off of x = 4csc(θ). It quickly dawned on me that I just needed to use the triangle to cross reference each answer based on the trig function the answer provided to see if the equation from each answer fit the triangle. It ended up being very simple.

The 9th question on the test was the last one that I had a hard time working through. Here it is:

Question 7

In retrospect, this question seems super straightforward. Since there were two functions, x and e^4x, I assumed it was an integration by parts question but when I set x as the antiderivative in the equation (x = a(x)) to use its derivative in the second step (x⁰ = 1 = a’(x)), I confused the fact that I was MULTIPLYING 1 by e^4x and not ADDING them together. For longer than I’d like to admit, I kept thinking that to integrate ∫ 1 * e^4x dx I was going to have to use the reverse power rule on 1, i.e. ∫ 1 dx = ∫ 1x⁰ dx = x + C, which is definitely wrong. It took me awhile to realize my mistake but I eventually figured it out (without cheating!). Afterwards I quickly finished the following 2 questions on the test to pass it with a perfect score.

This coming week I’ll be moving on to the unit Differential Equations (1,100/1,300 M.P.). There are 7 videos and 2 exercises that I need to get through before I can start the unit test. In a way I’m looking forward to revisiting this unit to get a stronger feel for differential equations. I’m hoping that I’ll find them easier to understand this time around than I did the first time, and I’m fairly optimistic that I will. Once I’m through Differential Equations, I’ll have 2 more units to get through (both of which are sitting at 0 M.P. and are huge) and then I’ll finally (FINALLY), after ~3 to 3.5 years of working on it, have gotten through calculus.

(Me once I finish Calc.2 ⬇️.)