Question 1

who's confused cos I'm definitely baffled o_0

Accepted Answer

At least you're introducing trigonometry into your communication, cos that's not easy.

Question 2

Repeatedly the idea of multiplying the dx out is being referred to as unorthodox. Is there a more formal or mainstream way of doing or thinking about this? It makes intuitive sense to me, but I'm wondering if there's something I should know about the "proper" method.

Accepted Answer

try googling "integration by parts"

Question 3

if trigonometry enters this field, am I gone?

Accepted Answer

Pretty vague question. Trigonometry is extensively integrated into calculus, so you'll eventually encounter it. That said, if you know your trig identities, it's really easy. Trust me, going forward, the easiest thing about integrals is doing the integral itself, be it trig, logs or anything else. Once you learn about double and triple integrals (yeah they exist), you'll see that the integration is the easiest part of it.

Question 4

Is there a video somewhere that goes over just u, du, dx and how they are chosen or derived?

Accepted Answer

u is just the variable that was chosen to represent what you replace.

du and dx are just parts of a derivative, where of course u is substituted part fo the function.  u will always be some function of x, so you take the derivative of u with respect to x, or in other words du/dx.

There should be videos on this playlist such as "u-substitution intro" and a few others I would watch, but you would also want to watch chain rule videos as well, though here is my explanation.   am assuming familiarity with the chain rule.

The chain rule starts with a composite function f(g(x)).  Such as sin(x^2), where one function is the sine operation and the other is the squared operation.  For the sake of clarity g(f(x)) would be sin^2(x).

Anyway, the chain rule says if you take the derivative with respect to x of f(g(x)) you get f'(g(x))*g'(x).  That means if you have a function in THAT form, you can take the integral of it to look like f(g(x)).  The process of doing this is traditionally u substitution.

So you start with f'(g(x))*g'(x).  the first step is to make u=g(x)  that way, when you take the derivative of u with respect to x (in other words du/dx) this gets you g'(x)  So now you know what g(x) is in f(g(x)).  since you start with f'(g(x))*g'(x) you ust have to take the integral of f'(g(x)) to get f(g(x)), though it's easier if g(x) is just a single variable, so we substitute in u for g(x).  of course at the end you need to re-substitute g(x) for it.

For notation's sake, and to ensure everything works out logically du/dx = g'(x) is changed to du = g'(x) dx because when you take the integral of something you add dx (as long as it is with respect to x.)  so the integral of f'(g(x))*g'(x) dx gets g'(x) dx replaced with du because f'(g(x)) becomes f'(u).  so now you have the integral of f'(u) du which of course becomes f(u), then you replace u with g(x) to get f(g(x)) effectively undoing the chain rule.

Let me know if this did not help.  And of course, I should mention with indefinite integrals you always put a + C at the end of your answer.

Question 5

If u substitution does not apply in ∫x^2 cos(2x)dx which approach should we use?

Accepted Answer

The product of a polynomial and a trig function is a classic example of integration by parts.

Question 6

Can I use u-substitution to find anti-derivative of (x^2 + 1)^2?   u = (x^2 + 1) and du/dx = 2x

Accepted Answer

That won't work. Replacing x²+1 by u leaves you with u²dx, but the du=2xdx part doesn't help you.

However, this is just a polynomial. You can expand it as x⁴+2x²+1, then use the power rule.

Question 7

can we multiply/divide by a variable?

Accepted Answer

As long as you do it to both sides of the equation, yes, you can (with the exception of the variable being = 0)

Question 8

So I have this question that has me baffled. It is: find the integral of *2sec^2⋅(x)tan(x)*

Now to solve it, I need u substitution. Well, when I saw the question, I found that I have two options for u, and both will work. Here's what I did:

*u = tan(x)*
du = sec^2(x) dx
dx = du/sec^2(x)

Putting this in the question, I get: 
int(2sec^2(x)⋅u du/sec^2(x)) = int(2u du)
= 2int(u du) = 2×1/2 u^2 = u^2 = tan^2(x) + C

So here we got the integral is *tan^2(x) + C*.

Now the second option

*u = sec(x)*
du = sec(x)tan(x) dx
dx = du/sec(x)tan(x)

Putting this in the question, *_but only substituting for one sec(x)_*, I get: 
int(2u⋅sec(x)tan(x) du/sec(x)tan(x)) = int(2u du) = u^2 (as done above) = sec^2(x) + C

Here, I get *sec^2(x) + C*!!

How's that possible??

Accepted Answer

Both are correct answers. The fun part about integrals is that you can have several antiderivatives.

If you want, differentiate both of them. For tan^(2)(x), you get 2tan(x)sec^(2)(x) = 2sec^(2)(x)tan(x). For sec^(2)(x), you get 2sec(x)sec(x)tan(x) = 2sec^(2)(x)tan(x). Same answer.

Question 9

Would you please help me to solve ∫6x^2e^2xdx?

Accepted Answer

I don't think you'll be able to solve this using 𝘶-substitution.
I used integration by parts (it took two iterations). There are several integration by parts videos on KA. I'd recommend you watch them all, but this one is closest to the problem you are trying to solve:
https://www.khanacademy.org/math/calculus-home/integration-techniques-calc/integration-by-parts-calc/v/integration-by-parts-twice-for-antiderivative-of-x-2-e-x

Integral Calculus

Course: Integral Calculus > Unit 1

𝘶-substitution

Using $u$ ‍-substitution with indefinite integrals

Common mistake: getting incorrect expressions for $u$ ‍ or $d u$ ‍

Common mistake: not realizing $u$ ‍-substitution is called for

Another common mistake: confusing the inner function and its derivative

Sometimes we need to multiply/divide the integral by a constant.

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