I once again got my ass kicked on KA this week. I didn’t get remotely close to finishing the Unit Test let alone get through it twice which is what I said I wanted to do at the end of my last post. I made it through a grand total of 8 questions and got 3 of them wrong… I don’t remember half of the things that are coming up which makes sense considering how long ago I started this unit. It’s definitely demoralizing but, even though it feels like I’m making zero headway on understanding integrals overall, I know that I’m making progress, it’s just going A LOT slower than I want it to and thought it would.
On Tuesday morning I got through questions 6 to 8 and then got stumped on question 9. The question asked me to evaluate the integral ∫ 8x/(√ 1 – 4x2) dx which I thought required me to use a trig substitution for the denominator. I spent half of Tuesday and all of Wednesday trying to solve this question and ended up getting it wrong. It turned out that I was supposed to use u-substitution all along:
Question 9:
(Note: In my eyes the denominator, √(1 – 4x2), looks a lot like √(c2 – a2) = b which is why I thought I needed to sub in something like √(1 – sin2(x)) = √(cos2(x)). I didn’t realize that I should have seen the denominator as u and the numerator as du.)
- 0∫1/4 8x/√(1 – 4x2) dx =
- b(x) = u = 1 – 4x2
- b’(x) = du = –8x
- = –1 * 0∫1/4 –1 * 8x/√(1 – 4x2) dx
- = –1 * 0∫1/4 –8x/√(1 – 4x2) dx
- = –1 * 0∫1/4 b’(x) * 1/√(b(x)) dx
- (Note: I know that when you use u-sub in a definite integral you have to change the values of the bounds, but I don’t really know how to do it so for now I usually just leave them and then replace u with the function it took the place of inside the integrand at the end.)
- = –1 * 0∫1/4 1/√u du
- = –1 * 0∫1/4 (u)-1/2 du
- = –1 * 2u1/2 o|1/4
- = –2u1/2 o|1/4
- = –2(1 – 4x2)1/2 o|1/4
- = (–2(1 – 4(1/4)2)1/2) – (–2(1 – 4(0)2)1/2)
- = (–2(1 – 4(1/16))1/2) – (–2(1 – 0)1/2)
- = (–2(1 – 4/16)1/2) – (–2(1)1/2)
- = (–2(4 – 1/4)1/2) – (–2(1))
- = (–2(3/4)1/2) – (–2)
- = (–2(√3/2)) – (–2)
- = (–√3) + 2
- = 2 – √3
Question 11:
(Note: There were 3 things I needed to do/understand in order to solve this question which were 1) know the derivative of arcsin(x), 2) know the derivative of arcsin(x/k), and 3) use the completing-the-square method inside the integral in order for the integral to take the shape of the derivative of arcsin(x/k).)
Step 1 – Understand the derivative of sin-1(x):
- d/dx[sin-1(x)] = (?)
- y = sin-1(x)
- sin(y) =
sin(sin-1(x))
- sin(y) = x
- d/dx[sin(y)] = d/dx[x]
- cos(y) * dy/dx = 1
- dy/dx = 1/cos(y)
- (Note: At this point you have to think about a right triangle where one of the angles is y. Since sin(y) = x = x/1 = Opp/Hyp, that means the opposite side of angle y is x and the hypotenuse is 1. Knowing that a2 + b2 = c2, if you say that a = sin(y) = x and c = hyp = 1, then you can state that b = √(1 – x2). Since b is the adjacent side of the triangle, then cos(y) = b = √(1 – x2) which is what you input into dy/dx = 1/cos(y).)
- dy/dx = 1/√(1 – x2)
- d/dx[sin-1(x)] = d/dx[y] = dy/dx = 1/√(1 – x2)
- d/dx[sin-1(x)] = 1/√(1 – x2)
Step 2 – Use the derivative of sin-1(x) to solve the derivative of sin-1(x/k):
- d/dx[sin-1(x/k)] = (?)
- (Note: It’s important to keep in mind is that 1/k is a constant.)
- a(x) = sin-1(x)
- a’(x) = 1/√(1 – x2)
- b(x) = x/k
- b’(x) = d/dx[x/k]
- = d/dx[ x * 1/k]
- = 1/k * d/dx[x]
- = 1/k * (1)
- = 1/k
- d/dx[a(b(x))] = a’(b(x)) * b’(x)
- = 1/√(1 – (x/k)2) * 1/k
- = 1/√(1 – (x2/k2) * 1/k
- = 1/√( k2/k2 – (x2/k2)) * 1/k
- = 1/√((k2 – x2)/k2) * 1/k
- = 1/√(k2 – x2)/√k2 * 1/k
- = 1/√(k2 – x2) * 1/1/√k2 * 1/k
- = 1/√(k2 – x2) * √k2 * 1/k
- = 1/√(k2 – x2) * k * 1/k
- = 1/√(k2 – x2) *
k * 1/k
- = 1/√(k2 – x2)
Step 3 – Using the derivative of sin-1(x/k), complete the square of ∫ 1/√( –x2 – 6x + 40) dx for it to take the shape of the derivative of sin-1(x/k) and then evaluate:
- ∫ 1/√( –x2 – 6x + 40) dx = ∫ 1/√( –1 * (x2 + 6x – 40)) dx
- = ∫ 1/√( –1 * (x2 + 6x + 9 – 9 – 40)) dx
- = ∫ 1/√( –1 * (x2 + 6x + 9 – 49)) dx
- = ∫ 1/√( –1 * ((x + 3)2 – 49)) dx
- = ∫ 1/√( –(x + 3)2 + 49) dx
- = ∫ 1/√( –(x + 3)2 + 72) dx
- = ∫ 1/√(72 – (x + 3)2) dx
- = ∫ 1/√(k2 – x2)
- k = 7
- x = (x + 3)
- = arcsin(x/k)
- = arcsin((x + 3)/7) + C
Question 13:
In this question I had no idea what was going on/what was being asked of me. I thought I had to evaluate the integral by inputting 9 into the upper-bound but that was wrong. As you can see from the screen shot, all I needed to do was take the integrand and input 9 into the variable t. I can’t remember how this works or what is happening here which is pretty sad and frustrating. 😔
Because I’m super cool, sometimes I’ll watch videos about math on YouTube at night and this week I started watching this video:
I’m only ~30 mins in so far but I’m hoping that watching more of this vid will help me wrap my head around some of the more tricky integrals. From what I’ve watched so far, this video has given me a bit of confidence that, even though I still feel pretty overwhelmed by integrals, I’m actually making progress on them and slowly starting to grasp how they work and the different methods used to solve them. Part of me feels like I’m getting close to understanding integrals but I really have no idea if that’s actually true so it’s definitely frustrating.
This coming week I’ll be happy if I can just make it through the rest of the 19 questions on the unit test. I’ll have to go back and redo the exercises of the questions I got/get wrong so I have a feeling I may not get through the unit test until the end of February. 😔 It’s a bit insane how long I’ve been working on the unit Integrals (2,560/3,200 M.P.) and I’m definitely more and more demoralized. But regardless, I’m going to keep grinding until I get through this!! 💪🏼 😤 I’m just hoping things will start to click sooner rather than later.