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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 6

Lesson 11: Integrating using substitution- 𝘶-substitution intro
- 𝘶-substitution: multiplying by a constant
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: defining 𝘶 (more examples)
- 𝘶-substitution
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: rational function
- 𝘶-substitution: logarithmic function
- 𝘶-substitution warmup
- 𝘶-substitution: indefinite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution with definite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution: definite integral of exponential function

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# 𝘶-substitution warmup

AP.CALC:

FUN‑6 (EU)

, FUN‑6.D (LO)

, FUN‑6.D.1 (EK)

Before diving into our practice exercise, gain some risk-free experience performing 𝘶-substitution.

**Find each indefinite integral.**

# Problem 1

# Problem 2

# Problem 3

# Problem 4

## Want to join the conversation?

- Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)(10 votes)
- 1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u).

This problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to.(16 votes)

- =∫1/u^2 du shoudn't be = ln(|u^2|)?(6 votes)
- You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.(4 votes)

- in problem 4 why is xdx= 3?(2 votes)
- if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx(3 votes)

- This is how I answered question 3:

int(e^(4x))dx

u = 4x

du = 4

int(e^(udu))dx

= int(e^(4x * 4))dx

= int(e^(16x))dx

= e^16x

How is my way wrong, and where did I go wrong?(2 votes)- First off, for u = 4x, du equals 4dx, not only 4.

Next, when you do a u-sub, everything becomes u. You cannot have any x's in it.

Now, onto the question. We have int(e^(4x)dx). Now, as du = 4dx, dx = du/4. And, e^(4x) = e^(u). Making these substitutions, we get int(e^(u)du/4). The 1/4 comes out and we have (1/4)int(e^(u)du). This gives (1/4)e^(u) + c. As u was 4x, on resubstituting, we get (1/4)e^(4x) + c. See if this made sense now.(2 votes)

- In problem 2, why the negative?

1/u^2 * du

-1/u(1 vote)- Reverse power rule.

∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.(3 votes)

- how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)(1 vote)
- ∫ 4x / sqrt(1 - x^4) dx =

2 ∫ 2x / sqrt(1 - (x^2)^2) dx

Let u = x^2, du = 2x dx, then

2 ∫ 2x dx / sqrt(1 - (x^2)^2) =

2 ∫ du / sqrt(1 - u^2) =

2 arcsin(u) + C =

2 arcsin(x^2) + C.

Hope that I helped.(3 votes)

- I was given the problem:
`∫ sin³(x)cos(x)dx = ? + C`

I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.(1 vote)- sin(x)^4/4 is correct however the exponent is in incorrect spot. The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4.

You need type sin^4(x)/4 or alternatively (sin(x))^4/4.(2 votes)

- In the first question, is it right to take cos(x^2) as u?(0 votes)
- If you choose cos(x^2) as your u, your du ends up being -sin(x^2)*2x*dx. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate.(4 votes)

- Actually the problem 3 can already be solved by using the integration formula of e.(1 vote)
- In order for most of these to work, the constant multiple rule must apply to integrals in exactly the same way that it applies to derivatives. Is that assumption correct?(1 vote)
- Yes the constant multiple rule applies for both derivatives and integrals(1 vote)