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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. :)(8 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.(13 votes)

- =∫1/u^2 du shoudn't be = ln(|u^2|)?(4 votes)
- You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.(3 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)

- 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)

- 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.(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)

- 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)

- =∫1/u^2 du = −1/u +C

Anyone could explain why appeared a negative signal on −1/u +C?(1 vote)- Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the curve at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 . This should hopefully provide some intuition for the negative sign.

For the rigorous proof, try finding d/dx (-1/x).(1 vote)