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AP Calculus AB Flashcards: Integrating Using Substitution

Study Integrating Using Substitution in AP Calculus AB with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Integrating Using Substitution, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus AB Flashcards: Integrating Using Substitution

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QUESTION

What substitution simplifies the integral of (2x+1)5(2x + 1)^5(2x+1)5?

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ANSWER

u=2x+1u = 2x + 1u=2x+1. Let u=2x+1u = 2x + 1u=2x+1 to transform (2x+1)5(2x + 1)^5(2x+1)5 into u5u^5u5 for easier integration.

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Flashcard 1: What substitution simplifies the integral of (2x+1)5(2x + 1)^5(2x+1)5?

Answer: u=2x+1u = 2x + 1u=2x+1. Let u=2x+1u = 2x + 1u=2x+1 to transform (2x+1)5(2x + 1)^5(2x+1)5 into u5u^5u5 for easier integration.

Flashcard 2: What is the substitution for uuu if du=−sin x dxdu = -\text{sin } x \, dxdu=−sin xdx?

Answer: u=cos xu = \text{cos } xu=cos x. Since ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin xdxd​(cosx)=−sinx, if du=−sin⁡xdxdu = -\sin x dxdu=−sinxdx, then u=cos⁡xu = \cos xu=cosx.

Flashcard 3: What is the integral of 1u du\frac{1}{u} \, duu1​du?

Answer: ln ∣u∣+C\text{ln } |u| + Cln ∣u∣+C. The antiderivative of 1u\frac{1}{u}u1​ is ln⁡∣u∣\ln|u|ln∣u∣ plus the constant of integration.

Flashcard 4: Determine dududu for u=sin xu = \text{sin } xu=sin x in substitution.

Answer: du=cos x dxdu = \text{cos } x \, dxdu=cos xdx. The derivative of u=sin⁡xu = \sin xu=sinx is cos⁡x\cos xcosx, so du=cos⁡xdxdu = \cos x dxdu=cosxdx.

Flashcard 5: Determine dududu if u=tan xu = \text{tan } xu=tan x in substitution.

Answer: du=sec2x dxdu = \text{sec}^2 x \, dxdu=sec2xdx. The derivative of u=tan⁡xu = \tan xu=tanx is sec⁡2x\sec^2 xsec2x, so du=sec⁡2xdxdu = \sec^2 x dxdu=sec2xdx.

Flashcard 6: What is the integral of un duu^n \, duundu?

Answer: un+1n+1+C\frac{u^{n+1}}{n+1} + Cn+1un+1​+C, n≠−1n \neq -1n=−1. Use the power rule for integration: increase exponent by 1 and divide.

Flashcard 7: What is the integral of sin u du\text{sin } u \, dusin udu?

Answer: −cos u+C-\text{cos } u + C−cos u+C. The antiderivative of sin⁡u\sin usinu is −cos⁡u-\cos u−cosu plus the constant of integration.

Flashcard 8: What substitution simplifies the integral of xcos(x2)x \text{cos}(x^2)xcos(x2)?

Answer: u=x2u = x^2u=x2. Let u=x2u = x^2u=x2 so that du=2xdxdu = 2x dxdu=2xdx matches the xxx factor in xcos⁡(x2)x\cos(x^2)xcos(x2).

Flashcard 9: Determine dududu if u=x2−1u = x^2 - 1u=x2−1 in substitution.

Answer: du=2x dxdu = 2x \, dxdu=2xdx. The derivative of u=x2−1u = x^2 - 1u=x2−1 is 2x2x2x, so du=2xdxdu = 2x dxdu=2xdx.

Flashcard 10: Find uuu for substitution if du=1x2 dxdu = \frac{1}{x^2} \, dxdu=x21​dx.

Answer: u=−1xu = -\frac{1}{x}u=−x1​. Since ddx(−1x)=1x2\frac{d}{dx}(-\frac{1}{x}) = \frac{1}{x^2}dxd​(−x1​)=x21​, if du=1x2dxdu = \frac{1}{x^2} dxdu=x21​dx, then u=−1xu = -\frac{1}{x}u=−x1​.

Flashcard 11: Identify uuu for substitution if du=2x dxdu = 2x \, dxdu=2xdx.

Answer: u=x2u = x^2u=x2. Since ddx(x2)=2x\frac{d}{dx}(x^2) = 2xdxd​(x2)=2x, if du=2xdxdu = 2x dxdu=2xdx, then u=x2u = x^2u=x2.

Flashcard 12: Find uuu for substitution if du=cos x dxdu = \text{cos } x \, dxdu=cos xdx.

Answer: u=sin xu = \text{sin } xu=sin x. Since ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos xdxd​(sinx)=cosx, if du=cos⁡xdxdu = \cos x dxdu=cosxdx, then u=sin⁡xu = \sin xu=sinx.

Flashcard 13: What is the integral of cos u du\text{cos } u \, ducos udu?

Answer: sin u+C\text{sin } u + Csin u+C. The antiderivative of cos⁡u\cos ucosu is sin⁡u\sin usinu plus the constant of integration.

Flashcard 14: Find uuu for substitution if du=7x6 dxdu = 7x^6 \, dxdu=7x6dx.

Answer: u=x7u = x^7u=x7. Since ddx(x7)=7x6\frac{d}{dx}(x^7) = 7x^6dxd​(x7)=7x6, if du=7x6dxdu = 7x^6 dxdu=7x6dx, then u=x7u = x^7u=x7.

Flashcard 15: Determine dududu if u=ln xu = \text{ln } xu=ln x in substitution.

Answer: du=1x dxdu = \frac{1}{x} \, dxdu=x1​dx. The derivative of u=ln⁡xu = \ln xu=lnx is 1x\frac{1}{x}x1​, so du=1xdxdu = \frac{1}{x} dxdu=x1​dx.

Flashcard 16: Determine dududu if u=3x4+2xu = 3x^4 + 2xu=3x4+2x in substitution.

Answer: du=(12x3+2) dxdu = (12x^3 + 2) \, dxdu=(12x3+2)dx. The derivative of u=3x4+2xu = 3x^4 + 2xu=3x4+2x is 12x3+212x^3 + 212x3+2, so du=(12x3+2)dxdu = (12x^3 + 2) dxdu=(12x3+2)dx.

Flashcard 17: What substitution simplifies the integral of xex2x e^{x^2}xex2?

Answer: u=x2u = x^2u=x2. Let u=x2u = x^2u=x2 so that du=2xdxdu = 2x dxdu=2xdx matches the xxx factor in xex2xe^{x^2}xex2.

Flashcard 18: Identify uuu for substitution if du=sec2x dxdu = \text{sec}^2 x \, dxdu=sec2xdx.

Answer: u=tan xu = \text{tan } xu=tan x. Since ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 xdxd​(tanx)=sec2x, if du=sec⁡2xdxdu = \sec^2 x dxdu=sec2xdx, then u=tan⁡xu = \tan xu=tanx.

Flashcard 19: Identify uuu for substitution if du=ex dxdu = e^x \, dxdu=exdx.

Answer: u=exu = e^xu=ex. Since ddx(ex)=ex\frac{d}{dx}(e^x) = e^xdxd​(ex)=ex, if du=exdxdu = e^x dxdu=exdx, then u=exu = e^xu=ex.

Flashcard 20: Determine dududu if u=2x3+1u = 2x^3 + 1u=2x3+1 in substitution.

Answer: du=6x2 dxdu = 6x^2 \, dxdu=6x2dx. The derivative of u=2x3+1u = 2x^3 + 1u=2x3+1 is 6x26x^26x2, so du=6x2dxdu = 6x^2 dxdu=6x2dx.

Flashcard 21: What is the integral of eu due^u \, dueudu?

Answer: eu+Ce^u + Ceu+C. The antiderivative of eue^ueu is eue^ueu plus the constant of integration.

Flashcard 22: Identify dududu if u=x3u = x^3u=x3 in substitution.

Answer: du=3x2 dxdu = 3x^2 \, dxdu=3x2dx. The derivative of u=x3u = x^3u=x3 is dudx=3x2\frac{du}{dx} = 3x^2dxdu​=3x2, so du=3x2dxdu = 3x^2 dxdu=3x2dx.

Flashcard 23: What substitution simplifies the integral of cos(x2)\text{cos}(x^2)cos(x2)?

Answer: u=x2u = x^2u=x2. Let u=x2u = x^2u=x2 to simplify the argument of the cosine function.

Flashcard 24: Find uuu for substitution if du=1x dxdu = \frac{1}{x} \, dxdu=x1​dx.

Answer: u=ln xu = \text{ln } xu=ln x. Since ddx(ln⁡x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}dxd​(lnx)=x1​, if du=1xdxdu = \frac{1}{x} dxdu=x1​dx, then u=ln⁡xu = \ln xu=lnx.

Flashcard 25: What substitution simplifies the integral of (2x+1)5(2x + 1)^5(2x+1)5?

Answer: u=2x+1u = 2x + 1u=2x+1. Let u=2x+1u = 2x + 1u=2x+1 to transform (2x+1)5(2x + 1)^5(2x+1)5 into u5u^5u5 for easier integration.

Flashcard 26: What is the substitution for uuu if du=−sin x dxdu = -\text{sin } x \, dxdu=−sin xdx?

Answer: u=cos xu = \text{cos } xu=cos x. Since ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin xdxd​(cosx)=−sinx, if du=−sin⁡xdxdu = -\sin x dxdu=−sinxdx, then u=cos⁡xu = \cos xu=cosx.

Flashcard 27: Identify uuu for substitution if du=5x4 dxdu = 5x^4 \, dxdu=5x4dx.

Answer: u=x5u = x^5u=x5. Since ddx(x5)=5x4\frac{d}{dx}(x^5) = 5x^4dxd​(x5)=5x4, if du=5x4dxdu = 5x^4 dxdu=5x4dx, then u=x5u = x^5u=x5.

Flashcard 28: Determine dududu if u=ln(4x)u = \text{ln}(4x)u=ln(4x) in substitution.

Answer: du=1x dxdu = \frac{1}{x} \, dxdu=x1​dx. The derivative of u=ln⁡(4x)u = \ln(4x)u=ln(4x) is 1x\frac{1}{x}x1​, so du=1xdxdu = \frac{1}{x} dxdu=x1​dx.

Flashcard 29: What is the integral of un duu^n \, duundu?

Answer: un+1n+1+C\frac{u^{n+1}}{n+1} + Cn+1un+1​+C, n≠−1n \neq -1n=−1. Use the power rule for integration: increase exponent by 1 and divide.

Flashcard 30: Determine dududu for u=sin xu = \text{sin } xu=sin x in substitution.

Answer: du=cos x dxdu = \text{cos } x \, dxdu=cos xdx. The derivative of u=sin⁡xu = \sin xu=sinx is cos⁡x\cos xcosx, so du=cos⁡xdxdu = \cos x dxdu=cosxdx.