v1 · padrão canônico

Lesson 84 — Technique: substitution (u-substitution)

Substitution u = g(x): inverse of the chain rule. The most widely used integration technique. Indefinite and definite versions. Pattern recognition.

Used in: 3.º ano do EM (17 anos) · Equiv. Math II japonês cap. 6 · Equiv. Klasse 12 alemã

\int f(g(x))\, g'(x)\, dx = \int f(u)\, du, \quad u = g(x)

Substituição: inversa da regra da cadeia. Quando o integrando contém o padrão f(g(x)) · g'(x), troque por u = g(x) e du = g'(x) dx. A integral nova deve ser mais simples.

Choose your door

Rigorous notation, full derivation, hypotheses

Teorema e procedimento

Teorema da mudança de variável

"A regra de substituição é o equivalente para integração da regra da cadeia para derivação." — OpenStax Calculus Vol. 1, §5.5

Demonstração. Se $F' = f$ , pela regra da cadeia: $(F \circ g)' = f(g(x)) \cdot g'(x)$ . Logo $F \circ g$ é antiderivada de $f(g(x)) g'(x)$ . Pelo TFC2:

$\int_a^b f(g(x)) g'(x)\, dx = F(g(b)) - F(g(a)) = \int_{g(a)}^{g(b)} f(u)\, du. \quad \square$

Procedimento mecânico

Sinal de que a substituição vai funcionar

O integrando deve conter $f(\text{algo})$ multiplicado pela derivada do "algo" (ou um múltiplo constante dessa derivada).

Exemplos de padrão:

Integrando	$u$	$du$
$2x\, e^{x^2}$	$x^2$	$2x\, dx$
$\cos x \cdot e^{\sin x}$	$\sin x$	$\cos x\, dx$
$\frac{x}{x^2 + 1}$	$x^2 + 1$	$2x\, dx$ (precisa ajuste $1/2$ )
$x^2(x^3 + 1)^5$	$x^3 + 1$	$3x^2\, dx$ (ajuste $1/3$ )

Exemplos resolvidos

Example— 2· Ajuste de constante (basico)

Problema. Calcule $\displaystyle\int x(x^2 + 5)^4\, dx$ .

Estratégia. $g(x) = x^2 + 5$ , $g'(x) = 2x$ . O integrando tem $x$ (metade do $g'$ ), então é necessário ajuste por $1/2$ .

Resolução. $u = x^2 + 5$ , $du = 2x\, dx$ , então $x\, dx = du/2$ .

$\int x(x^2+5)^4\, dx = \int u^4 \cdot \frac{du}{2} = \frac{1}{2} \cdot \frac{u^5}{5} + C = \frac{(x^2+5)^5}{10} + C.$

Verificação. $\left(\frac{(x^2+5)^5}{10}\right)' = \frac{5(x^2+5)^4 \cdot 2x}{10} = x(x^2+5)^4$ . Confere.

Fonte. APEX Calculus, §6.1, Exemplo 6.1.4 — CC-BY-NC.

Example— 3· Substituicao em integral definida com troca de limites (intermediario)

Problema. Calcule $\displaystyle\int_0^1 2x(x^2 + 1)^3\, dx$ .

Estratégia. $u = x^2 + 1$ , $du = 2x\, dx$ . Trocar os limites: $u(0) = 1$ , $u(1) = 2$ .

Resolução. $\int_0^1 2x(x^2+1)^3\, dx = \int_1^2 u^3\, du = \left[\frac{u^4}{4}\right]_1^2 = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}.$

Verificação. Sem troca de limites: antiderivada $(x^2+1)^4/4$ . Avaliar em $x=1$ e $x=0$ : $2^4/4 - 1^4/4 = 4 - 1/4 = 15/4$ . Confere.

Fonte. OpenStax Calculus Vol. 1, §5.5, Exemplo 5.38 — CC-BY-NC-SA.

Example— 4· Substituicao com função trigonometrica (intermediario)

Problema. Calcule $\displaystyle\int \sin^3 x \cos x\, dx$ .

Estratégia. $g(x) = \sin x$ , $g'(x) = \cos x$ — aparece explicitamente no integrando.

Resolução. $u = \sin x$ , $du = \cos x\, dx$ .

$\int \sin^3 x \cos x\, dx = \int u^3\, du = \frac{u^4}{4} + C = \frac{\sin^4 x}{4} + C.$

Verificação. $\left(\frac{\sin^4 x}{4}\right)' = \frac{4\sin^3 x \cos x}{4} = \sin^3 x \cos x$ . Confere.

Fonte. OpenStax Calculus Vol. 1, §5.5, Exemplo 5.36 — CC-BY-NC-SA.

Example— 5· Integral de tangente por substituicao (desafio)

Problema. Calcule $\displaystyle\int \tan x\, dx$ .

Estratégia. Escrever $\tan x = \sin x / \cos x$ , depois tomar $u = \cos x$ .

Resolução. $u = \cos x$ , $du = -\sin x\, dx$ , então $\sin x\, dx = -du$ .

$\int \frac{\sin x}{\cos x}\, dx = \int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C.$

Verificação. $(\ln|\sec x|)' = \frac{\sec x \tan x}{\sec x} = \tan x$ . Confere.

Fonte. APEX Calculus, §6.1, Exemplo 6.1.8 — CC-BY-NC.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 26Understanding 6Modeling 4Challenge 4

Ex. 84.1Application
Calcule $\int (x+1)^4\,dx$ usando $u=x+1$ .
Select the correct option
$\frac{(x+1)^5}{5} + C$ $(x+1)^5 + C$ $\frac{(x+1)^4}{4} + C$ $\frac{(x+1)^6}{6} + C$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 261
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Use $u=x+1$ , $du=dx$ : $\int u^4\,du = u^5/5+C = (x+1)^5/5+C$ . B esqueceu dividir por 5. C errou o expoente. D subiu demais o expoente.
Show step-by-step (with the why)
1. Escolha $u=x+1$ , $du=dx$ (sem fator de ajuste).
2. Substitua: $\int u^4\,du$ .
3. Integre: $u^5/5+C$ .
4. Volte: $(x+1)^5/5+C$ .
Ex. 84.2Application
Calcule $\int (x-1)^5\,dx$ usando $u=x-1$ .
Select the correct option
$\frac{(x-1)^6}{6} + C$ $(x-1)^6 + C$ $\frac{(x-1)^5}{5} + C$ $\frac{(x-1)^6}{5} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 262
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Use $u=x-1$ , $du=dx$ : $\int u^5\,du = u^6/6+C = (x-1)^6/6+C$ . B esqueceu dividir. C errou o expoente (ficou em 5). D errou o denominador.
Ex. 84.3Application
Calcule $\int (2x-3)^{-7}\,dx$ usando $u=2x-3$ .
Select the correct option
$-\frac{1}{12(2x-3)^6} + C$ $-\frac{1}{6(2x-3)^6} + C$ $\frac{1}{12(2x-3)^6} + C$ $-\frac{1}{7(2x-3)^6} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 263
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Use $u=2x-3$ , $du=2\,dx$ : $(1/2)\int u^{-7}\,du = (1/2)\cdot u^{-6}/(-6) = -1/(12u^6)+C = -1/(12(2x-3)^6)+C$ . B esqueceu o fator extra $1/2$ . C trocou o sinal. D errou o expoente.
Ex. 84.4ApplicationAnswer key
Calcule $\int (3x-2)^{-11}\,dx$ usando $u=3x-2$ .
Select the correct option
$-\frac{1}{30(3x-2)^{10}} + C$ $-\frac{1}{10(3x-2)^{10}} + C$ $\frac{1}{30(3x-2)^{10}} + C$ $-\frac{1}{33(3x-2)^{10}} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 264
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Use $u=3x-2$ , $du=3\,dx$ : $(1/3)\int u^{-11}\,du = (1/3)\cdot u^{-10}/(-10) = -1/(30(3x-2)^{10})+C$ . B esqueceu o fator $1/3$ . C trocou o sinal. D usou $3\times(-11)=-33$ erroneamente.
Show step-by-step (with the why)
1. $u=3x-2$ , $du=3\,dx$ , logo $dx=du/3$ .
2. Substitua: $(1/3)\int u^{-11}\,du$ .
3. Integre: $(1/3)\cdot u^{-10}/(-10) = -1/(30u^{10})$ .
4. Volte: $-1/(30(3x-2)^{10})+C$ .
Ex. 84.5Application
Calcule $\int \frac{x}{\sqrt{x^2+1}}\,dx$ usando $u=x^2+1$ .
Select the correct option
$\sqrt{x^2+1} + C$ $\frac{1}{2}\sqrt{x^2+1} + C$ $\ln|x^2+1| + C$ $\frac{x^2}{2}\sqrt{x^2+1} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 265
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Use $u=x^2+1$ , $du=2x\,dx$ : $(1/2)\int u^{-1/2}\,du = (1/2)\cdot 2u^{1/2}+C = \sqrt{x^2+1}+C$ . B esqueceu que $(1/2)\cdot 2=1$ . C usou antiderivada de $1/u$ . D não integrou.
Ex. 84.6Application
Calcule $\int x\sqrt{1-x^2}\,dx$ usando $u=1-x^2$ .
Select the correct option
$-\frac{1}{3}(1-x^2)^{3/2} + C$ $\frac{1}{3}(1-x^2)^{3/2} + C$ $-\frac{2}{3}(1-x^2)^{3/2} + C$ $(1-x^2)^{1/2} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 266
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Use $u=1-x^2$ , $du=-2x\,dx$ : $(-1/2)\int u^{1/2}\,du = (-1/2)\cdot(2/3)u^{3/2} = -u^{3/2}/3+C$ . B perdeu o sinal. C dobrou o fator. D não integrou corretamente.
Show step-by-step (with the why)
1. $u=1-x^2$ , $du=-2x\,dx$ , logo $x\,dx=-du/2$ .
2. Substitua: $\int x\sqrt{1-x^2}\,dx = (-1/2)\int u^{1/2}\,du$ .
3. Integre: $(-1/2)\cdot(2u^{3/2}/3) = -u^{3/2}/3$ .
4. Volte: $-(1-x^2)^{3/2}/3+C$ .
Ex. 84.7Understanding
Calcule $\int (x-1)(x^2-2x)^3\,dx$ usando $u=x^2-2x$ .
Select the correct option
$\frac{(x^2-2x)^4}{8} + C$ $\frac{(x^2-2x)^4}{4} + C$ $(x^2-2x)^3 + C$ $\frac{(x^2-2x)^4}{3} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 267
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Use $u=x^2-2x$ , $du=(2x-2)\,dx=2(x-1)\,dx$ : $(1/2)\int u^3\,du = u^4/8+C$ . B esqueceu o fator $1/2$ . C não integrou. D usou denominador errado.
Ex. 84.8ApplicationAnswer key
Calcule $\int (x^2-2x)(x^3-3x^2)^2\,dx$ usando $u=x^3-3x^2$ .
Select the correct option
$\frac{(x^3-3x^2)^3}{9} + C$ $\frac{(x^3-3x^2)^3}{3} + C$ $\frac{(x^3-3x^2)^2}{6} + C$ $3(x^2-2x)(x^3-3x^2)^2 + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 268
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Use $u=x^3-3x^2$ , $du=(3x^2-6x)\,dx=3(x^2-2x)\,dx$ : $(1/3)\int u^2\,du = u^3/9+C$ . B esqueceu o fator $1/3$ . C errou o expoente. D é a derivada, não a antiderivada.
Ex. 84.9Challenge
Calcule $\int \cos^3\theta\,d\theta$ usando $u=\sin\theta$ (dica: $\cos^2\theta=1-\sin^2\theta$ ).
Select the correct option
$\sin\theta - \frac{\sin^3\theta}{3} + C$ $\frac{\sin^3\theta}{3} + C$ $\cos\theta - \frac{\cos^3\theta}{3} + C$ $\frac{\cos^3\theta}{3} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 269
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Use $u=\sin\theta$ , $du=\cos\theta\,d\theta$ . Escreva $\cos^3\theta=\cos^2\theta\cdot\cos\theta=(1-\sin^2\theta)\cos\theta$ . Então $\int(1-u^2)\,du = u-u^3/3+C = \sin\theta-\sin^3\theta/3+C$ . B perdeu o primeiro termo. C e D usaram $u=\cos\theta$ (válido mas dá forma diferente).
Show step-by-step (with the why)
1. Escreva $\cos^3\theta = (1-\sin^2\theta)\cos\theta$ via identidade pitagórica.
2. Escolha $u=\sin\theta$ , $du=\cos\theta\,d\theta$ .
3. Substitua: $\int(1-u^2)\,du$ .
4. Integre: $u-u^3/3+C = \sin\theta-\sin^3\theta/3+C$ .
Ex. 84.10ChallengeAnswer key
Calcule $\int \sin^3\theta\,d\theta$ usando $u=\cos\theta$ (dica: $\sin^2\theta=1-\cos^2\theta$ ).
Select the correct option
$-\cos\theta + \frac{\cos^3\theta}{3} + C$ $\cos\theta - \frac{\cos^3\theta}{3} + C$ $\frac{\sin^3\theta}{3} - \sin\theta + C$ $-\frac{\cos^3\theta}{3} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 270
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Use $u=\cos\theta$ , $du=-\sin\theta\,d\theta$ . Escreva $\sin^3\theta=(1-\cos^2\theta)\sin\theta$ . Então $-\int(1-u^2)\,du = -(u-u^3/3)+C = -\cos\theta+\cos^3\theta/3+C$ . B perdeu o sinal global. C confundiu a escolha de $u$ . D perdeu o termo linear.
Ex. 84.11Challenge
Calcule $\int x(1-x)^{99}\,dx$ (use $u=1-x$ , escreva $x=1-u$ ).
Select the correct option
$-\frac{(1-x)^{101}}{101} + \frac{(1-x)^{100}}{100} + C$ $-\frac{(1-x)^{100}}{100} + C$ $\frac{(1-x)^{100}}{100} + C$ $x(1-x)^{100} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 271
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Use $u=1-x$ , $x=1-u$ , $dx=-du$ : $\int(1-u)u^{99}(-du)=-\int(u^{99}-u^{100})\,du=-u^{100}/100+u^{101}/101+C$ . B ignorou o fator $x=(1-u)$ . C trocou o sinal. D não integrou.
Show step-by-step (with the why)
1. $u=1-x$ , $x=1-u$ , $dx=-du$ .
2. Substitua: $\int(1-u)u^{99}(-du)=-\int u^{99}(1-u)\,du$ .
3. Expanda: $-\int(u^{99}-u^{100})\,du$ .
4. Integre: $-(u^{100}/100-u^{101}/101)+C=-u^{100}/100+u^{101}/101+C$ .
5. Volte: $-(1-x)^{100}/100+(1-x)^{101}/101+C$ .
Ex. 84.12Application
Calcule $\int t(1-t^2)^{10}\,dt$ .
Select the correct option
$-\frac{(1-t^2)^{11}}{22} + C$ $\frac{(1-t^2)^{11}}{11} + C$ $-\frac{(1-t^2)^{11}}{11} + C$ $\frac{(1-t^2)^{10}}{10} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 272
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Use $u=1-t^2$ , $du=-2t\,dt$ : $(-1/2)\int u^{10}\,du = (-1/2)\cdot u^{11}/11 = -(1-t^2)^{11}/22+C$ . B perdeu o sinal e o fator $1/2$ . C perdeu o fator $1/2$ . D não integrou.
Ex. 84.13ApplicationAnswer key
Calcule $\int (11x-7)^{-3}\,dx$ .
Select the correct option
$-\frac{1}{22(11x-7)^2} + C$ $-\frac{1}{2(11x-7)^2} + C$ $\frac{1}{22(11x-7)^2} + C$ $-\frac{1}{11(11x-7)^2} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 273
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Use $u=11x-7$ , $du=11\,dx$ : $(1/11)\int u^{-3}\,du=(1/11)\cdot u^{-2}/(-2)=-1/(22(11x-7)^2)+C$ . B esqueceu o fator $1/11$ . C trocou o sinal. D usou denominador errado.
Ex. 84.14Application
Calcule $\int (7x-11)^4\,dx$ .
Select the correct option
$\frac{(7x-11)^5}{35} + C$ $\frac{(7x-11)^5}{5} + C$ $7(7x-11)^4 + C$ $\frac{(7x-11)^5}{7} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 274
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Use $u=7x-11$ , $du=7\,dx$ : $(1/7)\int u^4\,du=(1/7)\cdot u^5/5 = u^5/35+C$ . B esqueceu o fator $1/7$ . C derivou. D usou $1/7$ mas não $1/5$ .
Ex. 84.15Application
Calcule $\int \cos^3\theta\sin\theta\,d\theta$ .
Select the correct option
$-\frac{\cos^4\theta}{4} + C$ $\frac{\cos^4\theta}{4} + C$ $-\frac{\cos^3\theta}{3} + C$ $\frac{\sin^4\theta}{4} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 275
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Use $u=\cos\theta$ , $du=-\sin\theta\,d\theta$ : $-\int u^3\,du = -u^4/4+C = -\cos^4\theta/4+C$ . B perdeu o sinal. C errou o expoente. D usou $u=\sin\theta$ .
Ex. 84.16Application
Calcule $\int \sin^7\theta\cos\theta\,d\theta$ .
Select the correct option
$\frac{\sin^8\theta}{8} + C$ $\frac{\sin^7\theta}{7} + C$ $7\sin^6\theta\cos\theta + C$ $\frac{\sin^8\theta}{7} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 276
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Use $u=\sin\theta$ , $du=\cos\theta\,d\theta$ : $\int u^7\,du = u^8/8+C = \sin^8\theta/8+C$ . B errou o expoente. C é a derivada. D misturou expoentes.
Ex. 84.17ApplicationAnswer key
Calcule $\int \cos^2(\pi t)\sin(\pi t)\,dt$ .
Select the correct option
$-\frac{\cos^3(\pi t)}{3\pi} + C$ $-\frac{\cos^3(\pi t)}{3} + C$ $\frac{\cos^3(\pi t)}{3\pi} + C$ $-\frac{\cos^2(\pi t)}{2\pi} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 277
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Use $u=\cos(\pi t)$ , $du=-\pi\sin(\pi t)\,dt$ : $(-1/\pi)\int u^2\,du = (-1/\pi)\cdot u^3/3 = -\cos^3(\pi t)/(3\pi)+C$ . B esqueceu dividir por $\pi$ . C trocou o sinal. D errou o expoente.
Show step-by-step (with the why)
1. $u=\cos(\pi t)$ , $du=-\pi\sin(\pi t)\,dt$ , logo $\sin(\pi t)\,dt=-du/\pi$ .
2. Substitua: $\int u^2(-du/\pi)=(-1/\pi)\int u^2\,du$ .
3. Integre: $(-1/\pi)\cdot u^3/3 = -u^3/(3\pi)+C$ .
4. Volte: $-\cos^3(\pi t)/(3\pi)+C$ .
Ex. 84.18Understanding
Calcule $\int \sin^2 x\cos^3 x\,dx$ (dica: $\cos^2 x=1-\sin^2 x$ ; use $u=\sin x$ ).
Select the correct option
$\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C$ $\frac{\sin^3 x}{3} + C$ $\frac{\sin^2 x\cos^3 x}{5} + C$ $-\frac{\sin^5 x}{5} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 278
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Use $u=\sin x$ , $du=\cos x\,dx$ . Escreva $\cos^3 x=(1-\sin^2 x)\cos x$ : $\int u^2(1-u^2)\,du=\int(u^2-u^4)\,du=u^3/3-u^5/5+C$ . B omitiu o fator $\cos^2 x$ . C errou a forma. D omitiu o primeiro termo.
Ex. 84.19UnderstandingAnswer key
Calcule $\int t\sin(t^2)\cos(t^2)\,dt$ .
Select the correct option
$\frac{\sin^2(t^2)}{4} + C$ $\sin^2(t^2) + C$ $-\frac{\cos^2(t^2)}{4} + C$ $\frac{t^2\sin(t^2)\cos(t^2)}{2} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 279
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Use $u=\sin(t^2)$ , $du=2t\cos(t^2)\,dt$ : integral contém $t\sin(t^2)\cos(t^2)\,dt=(1/2)\sin(t^2)\cdot 2t\cos(t^2)\,dt=(1/2)u\,du$ . Então $(1/2)\int u\,du=u^2/4+C=\sin^2(t^2)/4+C$ . B perdeu o fator $1/4$ . C errou a escolha de $u$ . D não integrou.
Ex. 84.20Application
Calcule $\int t^2\cos^2(t^3)\sin(t^3)\,dt$ .
Select the correct option
$-\frac{\cos^3(t^3)}{9} + C$ $-\frac{\cos^2(t^3)}{6} + C$ $\frac{\cos^3(t^3)}{9} + C$ $-\frac{t^2\cos^3(t^3)}{9} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 280
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Use $u=\cos(t^3)$ , $du=-3t^2\sin(t^3)\,dt$ : $(-1/3)\int u^2\,du=(-1/3)\cdot u^3/3=-\cos^3(t^3)/9+C$ . B errou o expoente. C trocou o sinal. D não integrou o $t^2$ .
Ex. 84.21Application
Calcule $\int x^2(x^3-3)^2\,dx$ .
Select the correct option
$\frac{(x^3-3)^3}{9} + C$ $\frac{(x^3-3)^3}{3} + C$ $x^2(x^3-3)^2 + C$ $\frac{(x^3-3)^2}{6} + C$
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Use $u=x^3-3$ , $du=3x^2\,dx$ : $(1/3)\int u^2\,du=(1/3)\cdot u^3/3=u^3/9+C=(x^3-3)^3/9+C$ . B esqueceu o fator $1/3$ . C é a derivada. D errou o expoente.
Ex. 84.22Modeling
Calcule $\int_0^1 x\sqrt{1-x^2}\,dx$ .
Select the correct option
$\frac{1}{3}$ $\frac{1}{2}$ $1$ $0$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 292
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Use $u=1-x^2$ , $du=-2x\,dx$ . Limites: $u(0)=1$ , $u(1)=0$ . $(-1/2)\int_1^0 u^{1/2}\,du=(1/2)\int_0^1 u^{1/2}\,du=(1/2)\cdot[u^{3/2}/(3/2)]_0^1=(1/2)\cdot(2/3)=1/3$ . B usou o fator errado. C não ajustou. D confundiu com função ímpar.
Show step-by-step (with the why)
1. $u=1-x^2$ , $du=-2x\,dx$ , $x\,dx=-du/2$ .
2. Limites: $u(0)=1$ , $u(1)=0$ .
3. $(-1/2)\int_1^0 u^{1/2}\,du=(1/2)\int_0^1 u^{1/2}\,du$ .
4. $=(1/2)\cdot[2u^{3/2}/3]_0^1=(1/2)\cdot(2/3)=1/3$ .
Ex. 84.23Application
Calcule $\int_0^1 \frac{x}{1+x^2}\,dx$ .
Select the correct option
$\frac{1}{2}\ln 2$ $\ln 2$ $\frac{\ln 2}{4}$ $1$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 293
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Use $u=1+x^2$ , $du=2x\,dx$ . Limites: $u(0)=1$ , $u(1)=2$ . $(1/2)\int_1^2 1/u\,du=(1/2)[\ln u]_1^2=(\ln 2)/2$ . B esqueceu o fator $1/2$ . C usou fator errado. D não obteve $\ln$ .
Ex. 84.24ModelingAnswer key
Calcule $\int_0^2 \frac{t}{\sqrt{5+t^2}}\,dt$ .
Select the correct option
$3 - \sqrt{5}$ $\sqrt{9} - \sqrt{5}$ $3$ $2$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 294
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Use $u=5+t^2$ , $du=2t\,dt$ . Limites: $u(0)=5$ , $u(2)=9$ . $(1/2)\int_5^9 u^{-1/2}\,du=(1/2)[2u^{1/2}]_5^9=[u^{1/2}]_5^9=3-\sqrt{5}$ . B confundiu $\sqrt{9}=3$ . C avaliou apenas em $u=9$ . D errou a antiderivada.
Ex. 84.25Application
Calcule $\int_0^1 \frac{t^2}{1+t^3}\,dt$ .
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$\frac{\ln 2}{3}$ $\ln 2$ $\frac{1}{3}$ $\frac{\ln 3}{3}$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 295
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Use $u=1+t^3$ , $du=3t^2\,dt$ . Limites: $u(0)=1$ , $u(1)=2$ . $(1/3)\int_1^2 1/u\,du=(1/3)[\ln u]_1^2=(\ln 2)/3$ . B esqueceu o fator $1/3$ . C não obteve $\ln$ . D errou os limites transformados.
Ex. 84.26Application
Calcule $\int_0^{\pi/4} \sec^2\theta\tan\theta\,d\theta$ .
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$\frac{1}{2}$ $1$ $\frac{\sqrt{2}}{4}$ $0$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 296
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Use $u=\tan\theta$ , $du=\sec^2\theta\,d\theta$ . Limites: $u(0)=0$ , $u(\pi/4)=1$ . $\int_0^1 u\,du=[u^2/2]_0^1=1/2$ . B dobrou. C errou a forma. D confundiu com integral de função ímpar.
Ex. 84.27Application
Calcule $\int_0^{\pi/4} \frac{\sin\theta}{\cos^4\theta}\,d\theta$ .
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$\frac{1}{4} - \frac{(\sqrt{2}/2)^4}{4}$ $\frac{1}{4}$ $\frac{\sqrt{2}}{4}$ $-\frac{1}{4}$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 297
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Use $u=\cos\theta$ , $du=-\sin\theta\,d\theta$ . Limites: $u(0)=1$ , $u(\pi/4)=\sqrt{2}/2$ . $-\int_1^{\sqrt{2}/2}u^{-4}\,du=\int_{\sqrt{2}/2}^1 u^{-4}\,du=[u^{-3}/(-3)]_{\sqrt{2}/2}^1$ . Simplificando: $[-1/(3\cos^3\theta)]_0^{\pi/4}=(1/4)-(\sqrt{2}/2)^4/4$ . B e D ignoraram os limites corretamente. C errou a forma.
Ex. 84.28Understanding
Use mudança de variável para mostrar que $\int_0^\pi \cos^2(2\theta)\sin(2\theta)\,d\theta = 0$ . Qual é o valor?
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$0$ — os valores de $u$ nos limites são iguais após a substituição $\pi/6$ $1/3$ $2/3$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 306
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Use $u=\cos^2(2\theta)$ , $du=-4\cos(2\theta)\sin(2\theta)\,d\theta$ . Limites: $u(0)=1$ e $u(\pi)=\cos^2(2\pi)=1$ . Logo $\int_1^1(\cdots)\,du=0$ . Os outros distractores erram os limites transformados.
Ex. 84.29UnderstandingAnswer key
Use mudança de variável para mostrar que $\int_0^{\sqrt{\pi}} t\cos(t^2)\sin(t^2)\,dt = 0$ . Qual é o valor?
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$0$ — os limites transformados coincidem $\pi^2/8$ $1/4$ $\pi/2$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 307
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Use $u=\sin(t^2)$ , $du=2t\cos(t^2)\,dt$ . Limites: $u(0)=\sin 0=0$ e $u(\pi)=\sin(\pi^2)\approx\sin(9{,}87)\ne 0$ . Alternativamente, use $v=t^2$ : limites ficam $0$ a $\pi^2$ , depois $u=\sin v$ : limites $0$ a $\sin(\pi^2)\approx-0{,}43$ . A integral fica $(1/4)\int_0^{\sin(\pi^2)}u\,du$ ; por simetria do enunciado, o exercício original usa $\sqrt{\pi}$ como limite superior, tornando os limites de $u$ iguais a 0, e o resultado é 0.
Ex. 84.30ApplicationAnswer key
Calcule $\int \frac{-2\sec(4x)\tan(4x)}{\sec^2(4x)}\,dx$ por substituição $u=\sec(4x)$ .
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$-\frac{1}{2\sec^2(4x)} + C$ $\frac{1}{2\sec^2(4x)} + C$ $-\frac{1}{\sec(4x)} + C$ $\ln|\sec(4x)| + C$
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Solve online active-calculus/single · §5.3 · ex. 1
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Simplifique: $\frac{\sec(4x)\tan(4x)}{\sec^2(4x)}=\frac{\tan(4x)}{\sec(4x)}=\sin(4x)$ . Então $\int(-2\sin(4x))\,dx=(-2)\cdot(-\cos(4x)/4)/1+C$ . Alternativamente, use $u=\sec(4x)$ , $du=4\sec(4x)\tan(4x)\,dx$ : $-2\int u^{-2}\cdot du/(4u)=(-1/2)\int u^{-2}\,du=-1/(2u)+C$ . B trocou o sinal. C errou a potência. D confundiu o integrando.
Ex. 84.31Application
Calcule $\int -6\csc^2(5x)e^{\cot(5x)}\,dx$ por substituição $u=\cot(5x)$ .
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$\frac{6}{5}e^{\cot(5x)} + C$ $-6e^{\cot(5x)} + C$ $6e^{\cot(5x)} + C$ $\frac{6}{5}\cot(5x)e^{\cot(5x)} + C$
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Solve online active-calculus/single · §5.3 · ex. 2
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Use $u=\cot(5x)$ , $du=-5\csc^2(5x)\,dx$ : $-6\int e^u\cdot(-du/5)=(6/5)\int e^u\,du=(6/5)e^u+C=(6/5)e^{\cot(5x)}+C$ . B perdeu o fator $1/5$ . C trocou o sinal. D não integrou.
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1. $u=\cot(5x)$ , $du=-5\csc^2(5x)\,dx$ , logo $\csc^2(5x)\,dx=-du/5$ .
2. Substitua: $-6\int e^u(-du/5)=(6/5)\int e^u\,du$ .
3. Integre: $(6/5)e^u+C$ .
4. Volte: $(6/5)e^{\cot(5x)}+C$ .
Ex. 84.32Application
Calcule $\int t^3(t^4-6)^2\,dt$ .
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$\frac{(t^4-6)^3}{12} + C$ $\frac{(t^4-6)^3}{4} + C$ $\frac{(t^4-6)^3}{3} + C$ $t^3(t^4-6)^2 + C$
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Solve online active-calculus/single · §5.3 · ex. 3
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Use $u=t^4-6$ , $du=4t^3\,dt$ : $(1/4)\int u^2\,du=(1/4)\cdot u^3/3=u^3/12+C=(t^4-6)^3/12+C$ . B esqueceu dividir por 3. C esqueceu o fator $1/4$ . D é a derivada.
Ex. 84.33Application
Calcule a antiderivada de $f(x)=9x^3\cos(x^4)$ .
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$\frac{9}{4}\sin(x^4) + C$ $9\sin(x^4) + C$ $\frac{9}{4}\cos(x^4) + C$ $9x^4\cos(x^4) + C$
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Solve online active-calculus/single · §5.3 · ex. 4
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Use $u=x^4$ , $du=4x^3\,dx$ : $9\int\cos u\cdot(du/4)=(9/4)\int\cos u\,du=(9/4)\sin u+C=(9/4)\sin(x^4)+C$ . B esqueceu o fator $1/4$ . C integrou como $\sin u$ → $-\cos u$ com sinal errado. D não integrou.
Ex. 84.34Application
Calcule $\int \frac{\ln^7 z}{z}\,dz$ .
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$\frac{\ln^8 z}{8} + C$ $\frac{\ln^7 z}{7} + C$ $\frac{7\ln^6 z}{z} + C$ $\frac{\ln^8 z}{7} + C$
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Solve online active-calculus/single · §5.3 · ex. 5
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Use $u=\ln z$ , $du=1/z\,dz$ : $\int u^7\,du=u^8/8+C=(\ln z)^8/8+C$ . B errou o expoente. C é a derivada. D misturou os expoentes.
Ex. 84.35ApplicationAnswer key
Calcule $\int \frac{e^{5x}}{3+e^{5x}}\,dx$ .
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$\frac{1}{5}\ln|3 + e^{5x}| + C$ $\ln|3 + e^{5x}| + C$ $\frac{e^{5x}}{5} + C$ $\frac{1}{5(3+e^{5x})} + C$
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Solve online active-calculus/single · §5.3 · ex. 6
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Use $u=3+e^{5x}$ , $du=5e^{5x}\,dx$ : $(1/5)\int 1/u\,du=(1/5)\ln|u|+C=(1/5)\ln|3+e^{5x}|+C$ . B esqueceu o fator $1/5$ . C não integrou o denominador. D aplicou regra de potência a $1/u$ .
Show step-by-step (with the why)
1. $u=3+e^{5x}$ , $du=5e^{5x}\,dx$ , logo $e^{5x}\,dx=du/5$ .
2. Substitua: $\int(e^{5x}/u)\,dx=(1/5)\int(1/u)\,du$ .
3. Integre: $(1/5)\ln|u|+C$ .
4. Volte: $(1/5)\ln|3+e^{5x}|+C$ .
Ex. 84.36Modeling
Calcule $\int_{3\pi}^{3\pi/2} e^{-\cos q}\cdot\sin q\,dq$ pelo TFC com substituição $u=-\cos q$ .
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$e^{-\cos(3\pi/2)} - e^{-\cos(3\pi)}$ $e^1 - e^0$ $e^{-1} - e^0$ $0$
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Solve online active-calculus/single · §5.3 · ex. 7
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Use $u=-\cos q$ , $du=\sin q\,dq$ . Limites: $q=3\pi\to u=-\cos(3\pi)=1$ , $q=3\pi/2\to u=-\cos(3\pi/2)=0$ . $\int_1^0 e^u\,du=[e^u]_1^0=e^0-e^1=1-e\approx-1{,}72$ . Expressão exata: $e^{-\cos(3\pi/2)}-e^{-\cos(3\pi)}=1-e$ . B confundiu os limites. C errou o sinal. D afirmou ser zero.
Ex. 84.37Understanding
Compare os dois métodos para $\int 2x^3(x^4+1)\,dx$ : (A) expandir o produto; (B) usar $w=x^4+1$ . As respostas são as mesmas?
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Ambas as antiderivadas são corretas; diferem apenas em uma constanteApenas o método de expansão é corretoApenas o método de substituição é corretoAs duas formas representam funções diferentes
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Solve online active-calculus/single · §5.3 · ex. 8
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Expandindo: $\int 2x^3(x^4+1)\,dx=\int(2x^7+2x^3)\,dx=x^8/4+x^4/2+C$ . Por substituição $w=x^4+1$ : $(1/2)\int w\,dw=w^2/4+C=(x^4+1)^2/4+C$ . Expandindo: $x^8/4+x^4/2+1/4+C$ . Diferença: constante $1/4$ absorvida em $C$ . Ambas corretas. B e C são falsas. D ignora que constantes de integração diferem.
Show step-by-step (with the why)
1. Expansão: $\int(2x^7+2x^3)\,dx=x^8/4+x^4/2+C_1$ .
2. Substituição $w=x^4+1$ : $(1/2)\int w\,dw=w^2/4+C_2=(x^4+1)^2/4+C_2$ .
3. Expanda a segunda: $x^8/4+x^4/2+1/4+C_2$ . Diferença $= 1/4$ : constante.
4. Conclusão: mesma antiderivada com constantes diferentes. Ambas corretas.
Ex. 84.38Application
Calcule $\int -2(1+\tan t)^5\sec^2 t\,dt$ .
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$-\frac{(1+\tan t)^6}{3} + C$ $\frac{(1+\tan t)^6}{6} + C$ $-\frac{(1+\tan t)^6}{6} + C$ $-2(1+\tan t)^5\sec^2 t + C$
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Solve online active-calculus/single · §5.3 · ex. 9
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Use $u=1+\tan t$ , $du=\sec^2 t\,dt$ : $-2\int u^5\,du=-2\cdot u^6/6=-u^6/3+C=-(1+\tan t)^6/3+C$ . B perdeu o sinal e usou $6$ em vez de $3$ . C perdeu o fator de $2$ . D não integrou.
Ex. 84.39Challenge
Calcule $\int 3\cot(x)\ln(\sin x)\,dx$ (use $v=\ln(\sin x)$ ).
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$\frac{3}{2}\ln^2(\sin x) + C$ $3\ln^2(\sin x) + C$ $3\ln(\sin x)\cos x + C$ $\frac{3}{2}\ln^2(\cos x) + C$
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Solve online active-calculus/single · §5.3 · ex. 10
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Use $v=\ln(\sin x)$ , $dv=\cos x/\sin x\,dx=\cot x\,dx$ : $3\int v\,dv=3v^2/2+C=(3/2)\ln^2(\sin x)+C$ . B esqueceu o fator $1/2$ . C não integrou. D confundiu o argumento.
Show step-by-step (with the why)
1. Note que $\cot x=\cos x/\sin x$ .
2. Escolha $v=\ln(\sin x)$ , $dv=(\cos x/\sin x)\,dx=\cot x\,dx$ .
3. Substitua: $3\int v\,dv=3v^2/2+C$ .
4. Volte: $(3/2)\ln^2(\sin x)+C$ .
Ex. 84.40Modeling
Calcule $\int x\sqrt{x-1}\,dx$ usando $u=x-1$ (escreva $x=u+1$ ).
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$\frac{2}{3}(x-1)^{3/2} + 2(x-1)^{1/2} + C$ $\frac{2}{5}x(x-1)^{3/2} + C$ $\frac{2}{3}(x-1)^{3/2} + C$ $2\sqrt{x-1} + C$
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Solve online active-calculus/single · §5.3 · ex. 12
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Use $u=x-1$ , $x=u+1$ , $dx=du$ : $\int(u+1)/\sqrt{u}\,du=\int(u^{1/2}+u^{-1/2})\,du=2u^{3/2}/3+2u^{1/2}+C$ . Volte: $2(x-1)^{3/2}/3+2(x-1)^{1/2}+C$ . B ignorou o fator $(u+1)$ . C omitiu o segundo termo. D integrou apenas $u^{-1/2}$ .
Show step-by-step (with the why)
1. $u=x-1$ , $x=u+1$ , $dx=du$ .
2. Substitua: $\int\frac{u+1}{\sqrt{u}}\,du=\int(u^{1/2}+u^{-1/2})\,du$ .
3. Integre: $2u^{3/2}/3+2u^{1/2}+C$ .
4. Volte: $2(x-1)^{3/2}/3+2\sqrt{x-1}+C$ .

Fontes

Active Calculus — Boelkins · §5.1 · CC-BY-NC-SA. Motivação da substituição como desfazer a regra da cadeia.
APEX Calculus — Hartman et al. · §6.1 · CC-BY-NC. Procedimento mecânico, ajuste de constante, integral definida com troca de limites.
OpenStax Calculus Volume 1 · §5.5 · CC-BY-NC-SA. Teorema formal, exemplos com radicais e trigonometria, exercícios de integral definida.