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### Course: AP®︎ Calculus BC (2017 edition) > Unit 9

Lesson 7: 𝘶-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

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. :)(12 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.(21 votes)

- =∫1/u^2 du shoudn't be = ln(|u^2|)?(11 votes)
- You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.(6 votes)

- For problem 2, I don't understand why the antiderivative is - 1/u + C, when we have an indefinite integral of 1/u^2 du? Shouldn't it be 1/U + C?(4 votes)
- Apply the reverse power rule. 1/u^2 is just u^(-2). So, this gives [u^(-2+1)]/(-2 + 1) which is u^(-1)/-1 which is -1/u(6 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(7 votes)

- This is very hard stuff(4 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.(5 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)

- how to identify which variable to qualify for u-substitution?(1 vote)
- You probably meant which
**function**qualifies for u-sub.

Identifying which function to take as 'u' simply comes with experience. Some integrals like sin(x)cos(x)dx have an easy u-substitution (u = sin(x) or cos(x)) as the 'u' and the derivative are explicitly given. Some like 1/sqrt(x - 9) require a trigonometric ratio to be 'u'. Some other questions make you come up with a completely (seemingly) irrelevant 'u' which actually simplifies the integral. Practice lots of problems and in no time, you'll be able to figure out which function to take as 'u'.(3 votes)

- For problem 4, how does x get a coefficient of 1/3 in the equation du=1/3xdx.(2 votes)
- Well, on differentiating (1/6)x^(2) + 1, using the power rule, we get (1/6)(2)(x), which is (1/3)x(1 vote)