AP Calculus AB Flashcards: Integrating Using Substitution

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.

Flashcard 1: Determine $du$ if $u = \text{ln } x$ in substitution.

Answer: $du = \frac{1}{x} \, dx$. The derivative of $u = \ln x$ is $\frac{1}{x}$, so $du = \frac{1}{x} dx$.

Flashcard 2: Determine $du$ if $u = \text{ln}(4x)$ in substitution.

Answer: $du = \frac{1}{x} \, dx$. The derivative of $u = \ln(4x)$ is $\frac{1}{x}$, so $du = \frac{1}{x} dx$.

Flashcard 3: What is the integral of $u^n \, du$?

Answer: $\frac{u^{n+1}}{n+1} + C$, $n \neq -1$. Use the power rule for integration: increase exponent by 1 and divide.

Flashcard 4: Determine $du$ if $u = \text{tan } x$ in substitution.

Answer: $du = \text{sec}^2 x \, dx$. The derivative of $u = \tan x$ is $\sec^2 x$, so $du = \sec^2 x dx$.

Flashcard 5: Identify $u$ for substitution if $du = 2x \, dx$.

Answer: $u = x^2$. Since $\frac{d}{dx}(x^2) = 2x$, if $du = 2x dx$, then $u = x^2$.

Flashcard 6: Identify $u$ for substitution if $du = \text{sec}^2 x \, dx$.

Answer: $u = \text{tan } x$. Since $\frac{d}{dx}(\tan x) = \sec^2 x$, if $du = \sec^2 x dx$, then $u = \tan x$.

Flashcard 7: Identify $u$ for substitution if $du = 4x^3 \, dx$.

Answer: $u = x^4$. Since $\frac{d}{dx}(x^4) = 4x^3$, if $du = 4x^3 dx$, then $u = x^4$.

Flashcard 8: What is the integral of $\frac{1}{u} \, du$?

Answer: $\text{ln } |u| + C$. The antiderivative of $\frac{1}{u}$ is $\ln|u|$ plus the constant of integration.

Flashcard 9: Determine $du$ if $u = 5x + 3$ in substitution.

Answer: $du = 5 \, dx$. The derivative of $u = 5x + 3$ is $5$, so $du = 5 dx$.

Flashcard 10: Determine $du$ if $u = \text{ln } x$ in substitution.

Answer: $du = \frac{1}{x} \, dx$. The derivative of $u = \ln x$ is $\frac{1}{x}$, so $du = \frac{1}{x} dx$.

Flashcard 11: Identify $u$ for substitution if $du = \text{sec}^2 x \, dx$.

Answer: $u = \text{tan } x$. Since $\frac{d}{dx}(\tan x) = \sec^2 x$, if $du = \sec^2 x dx$, then $u = \tan x$.

Flashcard 12: Find $u$ for substitution if $du = \text{cos } x \, dx$.

Answer: $u = \text{sin } x$. Since $\frac{d}{dx}(\sin x) = \cos x$, if $du = \cos x dx$, then $u = \sin x$.

Flashcard 13: What substitution simplifies the integral of $(2x + 1)^5$?

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

Flashcard 14: What is the integral of $e^u \, du$?

Answer: $e^u + C$. The antiderivative of $e^u$ is $e^u$ plus the constant of integration.

Flashcard 15: What is the integral of $\text{cos } u \, du$?

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

Flashcard 16: Find the integral of $\text{sec}^2 u \, du$.

Answer: $\text{tan } u + C$. The antiderivative of $\sec^2 u$ is $\tan u$ plus the constant of integration.

Flashcard 17: Determine $du$ if $u = 2x^3 + 1$ in substitution.

Answer: $du = 6x^2 \, dx$. The derivative of $u = 2x^3 + 1$ is $6x^2$, so $du = 6x^2 dx$.

Flashcard 18: Determine $du$ if $u = \text{ln}(4x)$ in substitution.

Answer: $du = \frac{1}{x} \, dx$. The derivative of $u = \ln(4x)$ is $\frac{1}{x}$, so $du = \frac{1}{x} dx$.

Flashcard 19: Identify $du$ if $u = x^3$ in substitution.

Answer: $du = 3x^2 \, dx$. The derivative of $u = x^3$ is $\frac{du}{dx} = 3x^2$, so $du = 3x^2 dx$.

Flashcard 20: What is the integral of $\frac{1}{u} \, du$?

Answer: $\text{ln } |u| + C$. The antiderivative of $\frac{1}{u}$ is $\ln|u|$ plus the constant of integration.

Flashcard 21: Determine $du$ for $u = \text{sin } x$ in substitution.

Answer: $du = \text{cos } x \, dx$. The derivative of $u = \sin x$ is $\cos x$, so $du = \cos x dx$.

Flashcard 22: Determine $du$ if $u = 3x^4 + 2x$ in substitution.

Answer: $du = (12x^3 + 2) \, dx$. The derivative of $u = 3x^4 + 2x$ is $12x^3 + 2$, so $du = (12x^3 + 2) dx$.

Flashcard 23: Find $u$ for substitution if $du = 7x^6 \, dx$.

Answer: $u = x^7$. Since $\frac{d}{dx}(x^7) = 7x^6$, if $du = 7x^6 dx$, then $u = x^7$.

Flashcard 24: What substitution simplifies the integral of $x \text{cos}(x^2)$?

Answer: $u = x^2$. Let $u = x^2$ so that $du = 2x dx$ matches the $x$ factor in $x\cos(x^2)$.

Flashcard 25: Find the integral of $\text{sec}^2 u \, du$.

Answer: $\text{tan } u + C$. The antiderivative of $\sec^2 u$ is $\tan u$ plus the constant of integration.

Flashcard 26: What substitution simplifies the integral of $(2x + 1)^5$?

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

Flashcard 27: Find $u$ for substitution if $du = 3x^2 dx$ in $\frac{1}{(x^3+1)^2}$.

Answer: $u = x^3 + 1$. If $du = 3x^2 dx$, then $u = x^3 + 1$ since $\frac{d}{dx}(x^3 + 1) = 3x^2$.

Flashcard 28: What substitution simplifies the integral of $x e^{x^2}$?

Answer: $u = x^2$. Let $u = x^2$ so that $du = 2x dx$ matches the $x$ factor in $xe^{x^2}$.

Flashcard 29: What is the integral of $e^u \, du$?

Answer: $e^u + C$. The antiderivative of $e^u$ is $e^u$ plus the constant of integration.

Flashcard 30: Identify $u$ for substitution if $du = 5x^4 \, dx$.

Answer: $u = x^5$. Since $\frac{d}{dx}(x^5) = 5x^4$, if $du = 5x^4 dx$, then $u = x^5$.

Flashcard 31: What is the integral of $\text{sin } u \, du$?

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

Flashcard 32: Find $u$ for substitution if $du = \frac{1}{x} \, dx$.

Answer: $u = \text{ln } x$. Since $\frac{d}{dx}(\ln x) = \frac{1}{x}$, if $du = \frac{1}{x} dx$, then $u = \ln x$.

Flashcard 33: Determine $du$ if $u = x^2 - 1$ in substitution.

Answer: $du = 2x \, dx$. The derivative of $u = x^2 - 1$ is $2x$, so $du = 2x dx$.

Flashcard 34: Find $u$ for substitution if $du = 7x^6 \, dx$.

Answer: $u = x^7$. Since $\frac{d}{dx}(x^7) = 7x^6$, if $du = 7x^6 dx$, then $u = x^7$.

Flashcard 35: What is the substitution for $u$ if $du = -\text{sin } x \, dx$?

Answer: $u = \text{cos } x$. Since $\frac{d}{dx}(\cos x) = -\sin x$, if $du = -\sin x dx$, then $u = \cos x$.

Flashcard 36: Find $u$ for substitution if $du = \frac{1}{x^2} \, dx$.

Answer: $u = -\frac{1}{x}$. Since $\frac{d}{dx}(-\frac{1}{x}) = \frac{1}{x^2}$, if $du = \frac{1}{x^2} dx$, then $u = -\frac{1}{x}$.

Flashcard 37: Determine $du$ for $u = \text{sin } x$ in substitution.

Answer: $du = \text{cos } x \, dx$. The derivative of $u = \sin x$ is $\cos x$, so $du = \cos x dx$.

Flashcard 38: What substitution simplifies the integral of $\text{cos}(x^2)$?

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

Flashcard 39: Find $u$ for substitution if $du = \frac{1}{x} \, dx$.

Answer: $u = \text{ln } x$. Since $\frac{d}{dx}(\ln x) = \frac{1}{x}$, if $du = \frac{1}{x} dx$, then $u = \ln x$.

Flashcard 40: Determine $du$ if $u = 3x^4 + 2x$ in substitution.

Answer: $du = (12x^3 + 2) \, dx$. The derivative of $u = 3x^4 + 2x$ is $12x^3 + 2$, so $du = (12x^3 + 2) dx$.

Flashcard 41: Identify $u$ for substitution if $du = 5x^4 \, dx$.

Answer: $u = x^5$. Since $\frac{d}{dx}(x^5) = 5x^4$, if $du = 5x^4 dx$, then $u = x^5$.

Flashcard 42: Find $u$ for substitution if $du = 3x^2 dx$ in $\frac{1}{(x^3+1)^2}$.

Answer: $u = x^3 + 1$. If $du = 3x^2 dx$, then $u = x^3 + 1$ since $\frac{d}{dx}(x^3 + 1) = 3x^2$.

Flashcard 43: What is the integral of $u^n \, du$?

Answer: $\frac{u^{n+1}}{n+1} + C$, $n \neq -1$. Use the power rule for integration: increase exponent by 1 and divide.

Flashcard 44: What is the integral of $\text{cos } u \, du$?

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

Flashcard 45: Identify $u$ for substitution if $du = e^x \, dx$.

Answer: $u = e^x$. Since $\frac{d}{dx}(e^x) = e^x$, if $du = e^x dx$, then $u = e^x$.

Flashcard 46: Identify $du$ if $u = x^3$ in substitution.

Answer: $du = 3x^2 \, dx$. The derivative of $u = x^3$ is $\frac{du}{dx} = 3x^2$, so $du = 3x^2 dx$.

Flashcard 47: Determine $du$ if $u = x^2 - 1$ in substitution.

Answer: $du = 2x \, dx$. The derivative of $u = x^2 - 1$ is $2x$, so $du = 2x dx$.

Flashcard 48: Identify $u$ for substitution if $du = 4x^3 \, dx$.

Answer: $u = x^4$. Since $\frac{d}{dx}(x^4) = 4x^3$, if $du = 4x^3 dx$, then $u = x^4$.

Flashcard 49: Determine $du$ if $u = 5x + 3$ in substitution.

Answer: $du = 5 \, dx$. The derivative of $u = 5x + 3$ is $5$, so $du = 5 dx$.

Flashcard 50: What substitution simplifies the integral of $x \text{cos}(x^2)$?

Answer: $u = x^2$. Let $u = x^2$ so that $du = 2x dx$ matches the $x$ factor in $x\cos(x^2)$.

Flashcard 51: What is the integral of $\text{sin } u \, du$?

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

Flashcard 52: Identify $u$ for substitution if $du = e^x \, dx$.

Answer: $u = e^x$. Since $\frac{d}{dx}(e^x) = e^x$, if $du = e^x dx$, then $u = e^x$.

Flashcard 53: What substitution simplifies the integral of $x e^{x^2}$?

Answer: $u = x^2$. Let $u = x^2$ so that $du = 2x dx$ matches the $x$ factor in $xe^{x^2}$.

Flashcard 54: What is the substitution for $u$ if $du = -\text{sin } x \, dx$?

Answer: $u = \text{cos } x$. Since $\frac{d}{dx}(\cos x) = -\sin x$, if $du = -\sin x dx$, then $u = \cos x$.

Flashcard 55: What substitution simplifies the integral of $\cos(x^2)$?

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

Flashcard 56: Identify $u$ for substitution if $du = 2x \, dx$.

Answer: $u = x^2$. Since $\frac{d}{dx}(x^2) = 2x$, if $du = 2x dx$, then $u = x^2$.

Flashcard 57: Determine $du$ if $u = 2x^3 + 1$ in substitution.

Answer: $du = 6x^2 \, dx$. The derivative of $u = 2x^3 + 1$ is $6x^2$, so $du = 6x^2 dx$.

Flashcard 58: Find $u$ for substitution if $du = \text{cos } x \, dx$.

Answer: $u = \text{sin } x$. Since $\frac{d}{dx}(\sin x) = \cos x$, if $du = \cos x dx$, then $u = \sin x$.

Flashcard 59: Determine $du$ if $u = \text{tan } x$ in substitution.

Answer: $du = \text{sec}^2 x \, dx$. The derivative of $u = \tan x$ is $\sec^2 x$, so $du = \sec^2 x dx$.

Flashcard 60: Find $u$ for substitution if $du = \frac{1}{x^2} \, dx$.

Answer: $u = -\frac{1}{x}$. Since $\frac{d}{dx}(-\frac{1}{x}) = \frac{1}{x^2}$, if $du = \frac{1}{x^2} dx$, then $u = -\frac{1}{x}$.