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. :)

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.

=∫1/u^2 du shoudn't be = ln(|u^2|)?

You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.

in problem 4 why is xdx= 3?

if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx

In problem 2, why the negative? 1/u^2 * du -1/u

Reverse power rule. ∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.

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.

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.

how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)

∫ 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.

In the first question, is it right to take cos(x^2) as u?

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.

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𝘶-substitution warmup

AP.CALC:

FUN‑6 (EU)

FUN‑6.D (LO)

FUN‑6.D.1 (EK)

Google Classroom

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

Find each indefinite integral.

Problem 1

integral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals

plus, space, C

[Show me the solution, please!]

Problem 2

integral, start fraction, 3, x, squared, divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals

plus, space, C

[Show me the solution, please!]

Problem 3

integral, e, start superscript, 4, x, end superscript, d, x, equals

plus, space, C

[Show me the solution, please!]

Problem 4

integral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction, x, squared, plus, 1, end square root, d, x, equals

plus, space, C

[Show me the solution, please!]

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Roberto Dias
Posted 6 years ago. Direct link to Roberto Dias's post “Well, in the problem #2 w...”
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. :)
Button navigates to signup pageComment on Roberto Dias's post “Well, in the problem #2 w...”
(8 votes)
Answer
- Brielle of Earth
  Posted 5 years ago. Direct link to Brielle of Earth's post “1/(u^2) == u^(-2). When y...”
  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.
  Button navigates to signup page
  (13 votes)
Vítor Watanabe
Posted 3 years ago. Direct link to Vítor Watanabe's post “=∫1/u^2 du shoudn't be =...”
=∫1/u^2 du shoudn't be = ln(|u^2|)?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- cossine
  Posted 3 years ago. Direct link to cossine's post “You are reversing the pow...”
  You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.
  Button navigates to signup page
  (3 votes)
Najee Hillman
Posted 4 years ago. Direct link to Najee Hillman's post “in problem 4 why is xdx= ...”
in problem 4 why is xdx= 3?
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(2 votes)
Answer
- Lee
  Posted 4 years ago. Direct link to Lee's post “if du = 1/3xdx, you just ...”
  if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx
  Button navigates to signup page
  (3 votes)
Avalon Z
Posted 4 years ago. Direct link to Avalon Z's post “In problem 2, why the neg...”
In problem 2, why the negative?

1/u^2 * du
-1/u
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(1 vote)
Answer
- Alex
  Posted 4 years ago. Direct link to Alex's post “Reverse power rule. ∫ u^(...”
  Reverse power rule.
  ∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.
  Button navigates to signup page
  (3 votes)
scott.d.corwin
Posted 3 years ago. Direct link to scott.d.corwin's post “I was given the problem: ...”
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.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- cossine
  Posted 3 years ago. Direct link to cossine's post “sin(x)^4/4 is correct how...”
  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.
  Button navigates to signup page
  (3 votes)
Laurie Eyre
Posted 4 years ago. Direct link to Laurie Eyre's post “how to us u-substitution ...”
how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Alex
  Posted 4 years ago. Direct link to Alex's post “∫ 4x / sqrt(1 - x^4) dx =...”
  ∫ 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.
  Comment on Alex's post “∫ 4x / sqrt(1 - x^4) dx =...”
  (3 votes)
Nischal Chandur
Posted 6 years ago. Direct link to Nischal Chandur's post “In the first question, is...”
In the first question, is it right to take cos(x^2) as u?
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(0 votes)
Answer
- Samuli Niemi
  Posted 6 years ago. Direct link to Samuli Niemi's post “If you choose cos(x^2) as...”
  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.
  Button navigates to signup page
  (4 votes)
RussellSH519
Posted 2 years ago. Direct link to RussellSH519's post “Actually the problem 3 ca...”
Actually the problem 3 can already be solved by using the integration formula of e.
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(1 vote)
Answer
Avi Perl
Posted 3 years ago. Direct link to Avi Perl's post “In order for most of thes...”
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?
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(1 vote)
Answer
- Angell, J
  Posted 3 years ago. Direct link to Angell, J's post “Yes the constant multiple...”
  Yes the constant multiple rule applies for both derivatives and integrals
  Button navigates to signup page
  (1 vote)
Vítor Watanabe
Posted 3 years ago. Direct link to Vítor Watanabe's post “=∫1/u^2 du = −1/u +C A...”
=∫1/u^2 du = −1/u +C

Anyone could explain why appeared a negative signal on −1/u +C?
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(1 vote)
Answer
- The #1 Pokemon Proponent
  Posted 3 years ago. Direct link to The #1 Pokemon Proponent's post “Think about it this way. ...”
  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).
  Button navigates to signup page
  (1 vote)

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 6

𝘶-substitution warmup

Problem 1

Problem 2

Problem 3

Problem 4

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