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Lição 35 — Técnica de Substituição

Substituição u-du como inversa da Regra da Cadeia. Aplicações em integrais indefinidas e definidas (ajuste de limites). Estratégias para identificar a substituição adequada.

Used in: Cálculo 1 — Unidade 4 · USP MAC0105 · ITA MA-011

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

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Rigorous notation, full derivation, hypotheses

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Exemplos

Exercícios

Exercise list

30 exercises · 7 with worked solution (25%)

Application 26Understanding 2Modeling 1Challenge 1

Ex. 35.1UnderstandingAnswer key
Por que a substituição $u$ - $du$ é chamada de "mudança de variável"?
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A substituição $u = g(x)$ transforma a integral pela inversão da regra da cadeiaA substituição é um método para derivar funções compostasA substituição só funciona quando o integrando é um polinômioA substituição $u$ - $du$ é equivalente à integração por partes
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 254
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A substituição $u = g(x)$ reverte a regra da cadeia: se $F'=f$ , então $\frac{d}{dx}F(g(x))=f(g(x))g'(x)$ , logo $\int f(g(x))g'(x)\,dx = F(g(x))+C$ .
Ex. 35.2Understanding
Se $f = g \circ h$ , ao reverter a regra da cadeia $\dfrac{d}{dx}(g \circ h)(x) = g'(h(x))\,h'(x)$ , qual deve ser a escolha de $u$ ?
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$u = h(x)$ , a função interna da composição $u = g(x)$ , a função externa da composição $u = f(x)$ , a função a integrar $u = g'(x)$ , a derivada da função externa
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 255
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Na composição $f = g \circ h$ , a regra da cadeia dá $(g\circ h)'(x) = g'(h(x))\,h'(x)$ . Para reverter isso, fazemos $u = h(x)$ (função interna), pois $du = h'(x)\,dx$ cancela o fator extra.
Ex. 35.3ApplicationAnswer key
Calcule $\int (x+1)^4\,dx$ com a substituição $u = x+1$ .
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$\dfrac{(x+1)^5}{5} + C$ $\dfrac{(x+1)^5}{4} + C$ $(x+1)^5 + C$ $\dfrac{x^5}{5} + x^4 + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 261
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Faça $u = x+1$ , $du = dx$ . Então $\int u^4\,du = \frac{u^5}{5}+C = \frac{(x+1)^5}{5}+C$ .
Show step-by-step (with the why)
1. Escolha $u = x+1$ , logo $du = dx$ .
2. Substitua: $\int (x+1)^4\,dx = \int u^4\,du$ .
3. Integre: $\frac{u^5}{5}+C$ .
4. Reverta: $\frac{(x+1)^5}{5}+C$ .
Ex. 35.4ApplicationAnswer key
Calcule $\int (x-1)^5\,dx$ com a substituição $u = x-1$ .
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$\dfrac{(x-1)^6}{6} + C$ $\dfrac{(x-1)^6}{5} + C$ $\dfrac{x^6 - 6x^5}{6} + C$ $(x-1)^6 + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 262
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Faça $u = x-1$ , $du = dx$ : $\int u^5\,du = \frac{u^6}{6}+C = \frac{(x-1)^6}{6}+C$ .
Ex. 35.5Application
Calcule $\int (2x-3)^{-7}\,dx$ com a substituição $u = 2x-3$ .
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$-\dfrac{1}{12(2x-3)^6} + C$ $-\dfrac{1}{6(2x-3)^6} + C$ $\dfrac{1}{12(2x-3)^6} + C$ $-\dfrac{1}{12(2x-3)^{-8}} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 263
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Faça $u = 2x-3$ , $du = 2\,dx$ , logo $dx = du/2$ . Então $\int u^{-7}\frac{du}{2} = \frac{1}{2}\cdot\frac{u^{-6}}{-6}+C = -\frac{1}{12(2x-3)^6}+C$ .
Ex. 35.6ApplicationAnswer key
Calcule $\int (3x-2)^{-11}\,dx$ com a substituição $u = 3x-2$ .
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$-\dfrac{1}{30(3x-2)^{10}} + C$ $-\dfrac{1}{10(3x-2)^{10}} + C$ $\dfrac{1}{30(3x-2)^{10}} + C$ $-\dfrac{1}{3(3x-2)^{-10}} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 264
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Faça $u = 3x-2$ , $du = 3\,dx$ : $\int u^{-11}\frac{du}{3} = \frac{1}{3}\cdot\frac{u^{-10}}{-10}+C = -\frac{1}{30(3x-2)^{10}}+C$ .
Ex. 35.7Application
Calcule $\int \frac{x}{\sqrt{x^2+1}}\,dx$ com a substituição $u = x^2+1$ .
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$\sqrt{x^2+1} + C$ $\dfrac{1}{2}\sqrt{x^2+1} + C$ $2\sqrt{x^2+1} + C$ $\dfrac{x^2}{2}\sqrt{x^2+1} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 265
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Faça $u = x^2+1$ , $du = 2x\,dx$ , logo $x\,dx = du/2$ : $\int \frac{1}{\sqrt{u}}\frac{du}{2} = \sqrt{u}+C = \sqrt{x^2+1}+C$ .
Show step-by-step (with the why)
1. Faça $u = x^2+1$ , $du = 2x\,dx$ .
2. Escreva $x\,dx = du/2$ e substitua: $\int \frac{x}{\sqrt{x^2+1}}\,dx = \int \frac{du/2}{\sqrt{u}}$ .
3. Integre: $\frac{1}{2}\cdot 2\sqrt{u}+C = \sqrt{u}+C$ .
4. Reverta: $\sqrt{x^2+1}+C$ .
Ex. 35.8Application
Calcule $\int \frac{x}{ \sqrt{1-x^2} }\,dx$ com a substituição $u = 1-x^2$ .
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$-\sqrt{1-x^2} + C$ $\sqrt{1-x^2} + C$ $-\dfrac{1}{2}\sqrt{1-x^2} + C$ $\dfrac{x^2}{\sqrt{1-x^2}} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 266
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Faça $u = 1-x^2$ , $du = -2x\,dx$ : $\int \frac{x}{\sqrt{1-x^2}}\,dx = \int \frac{-du/2}{\sqrt{u}} = -\sqrt{u}+C = -\sqrt{1-x^2}+C$ .
Ex. 35.9Application
Calcule $\int (x-1)(x^2-2x)^3\,dx$ com a substituição $u = x^2-2x$ .
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$\dfrac{(x^2-2x)^4}{8} + C$ $\dfrac{(x^2-2x)^4}{4} + C$ $\dfrac{(x-1)(x^2-2x)^4}{4} + C$ $\dfrac{(x^2-2x)^3}{4} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 267
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Faça $u = x^2-2x$ , $du = (2x-2)\,dx = 2(x-1)\,dx$ , logo $(x-1)\,dx = du/2$ : $\int u^3\frac{du}{2} = \frac{u^4}{8}+C = \frac{(x^2-2x)^4}{8}+C$ .
Ex. 35.10Application
Calcule $\int (x^2-2x)(x^3-3x^2)^2\,dx$ com a substituição $u = x^3-3x^2$ .
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$\dfrac{(x^3-3x^2)^3}{9} + C$ $\dfrac{(x^3-3x^2)^3}{3} + C$ $\dfrac{(x^2-2x)(x^3-3x^2)^3}{3} + C$ $\dfrac{(x^3-3x^2)^2}{6} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 268
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Faça $u = x^3-3x^2$ , $du = (3x^2-6x)\,dx = 3(x^2-2x)\,dx$ : $\int u^2\frac{du}{3} = \frac{u^3}{9}+C = \frac{(x^3-3x^2)^3}{9}+C$ .
Ex. 35.11Application
Calcule $\int \cos^3\theta\,d\theta$ com a substituição $u = \sin\theta$ e a identidade $\cos^2\theta = 1-\sin^2\theta$ .
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$\sin\theta - \dfrac{\sin^3\theta}{3} + C$ $\sin\theta + \dfrac{\sin^3\theta}{3} + C$ $\cos\theta - \dfrac{\cos^3\theta}{3} + C$ $\dfrac{\sin^4\theta}{4} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 269
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Use $\cos^2\theta = 1-\sin^2\theta$ . Faça $u = \sin\theta$ , $du = \cos\theta\,d\theta$ : $\int\cos^3\theta\,d\theta = \int(1-u^2)\,du = u - \frac{u^3}{3}+C = \sin\theta - \frac{\sin^3\theta}{3}+C$ .
Show step-by-step (with the why)
1. Escreva $\cos^3\theta = \cos^2\theta\cdot\cos\theta = (1-\sin^2\theta)\cos\theta$ .
2. Faça $u = \sin\theta$ , $du = \cos\theta\,d\theta$ .
3. Substitua: $\int (1-u^2)\,du$ .
4. Integre: $u - u^3/3 + C$ .
5. Reverta: $\sin\theta - \sin^3\theta/3 + C$ .
Ex. 35.12Application
Calcule $\int \sin^3\theta\,d\theta$ com a substituição $u = \cos\theta$ e a identidade $\sin^2\theta = 1-\cos^2\theta$ .
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$-\cos\theta + \dfrac{\cos^3\theta}{3} + C$ $\cos\theta - \dfrac{\cos^3\theta}{3} + C$ $-\sin\theta + \dfrac{\sin^3\theta}{3} + C$ $\dfrac{\cos^4\theta}{4} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 270
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Use $\sin^2\theta = 1-\cos^2\theta$ . Faça $u = \cos\theta$ , $du = -\sin\theta\,d\theta$ : $\int\sin^3\theta\,d\theta = \int(1-u^2)(-du) = -u+\frac{u^3}{3}+C = -\cos\theta+\frac{\cos^3\theta}{3}+C$ .
Ex. 35.13Application
Calcule $\int \cos^3\theta\sin\theta\,d\theta$ .
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$-\dfrac{\cos^4\theta}{4} + C$ $\dfrac{\cos^4\theta}{4} + C$ $-\dfrac{\cos^3\theta}{3} + C$ $\dfrac{\sin^4\theta}{4} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 275
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Faça $u = \cos\theta$ , $du = -\sin\theta\,d\theta$ : $\int\cos^3\theta\sin\theta\,d\theta = \int u^3(-du) = -\frac{u^4}{4}+C = -\frac{\cos^4\theta}{4}+C$ .
Ex. 35.14Application
Calcule $\int \sin^7\theta\cos\theta\,d\theta$ .
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$\dfrac{\sin^8\theta}{8} + C$ $\dfrac{\sin^7\theta}{7} + C$ $8\sin^7\theta\cos\theta + C$ $\dfrac{\cos^8\theta}{8} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 276
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Faça $u = \sin\theta$ , $du = \cos\theta\,d\theta$ : $\int u^7\,du = \frac{u^8}{8}+C = \frac{\sin^8\theta}{8}+C$ .
Ex. 35.15Application
Calcule $\int \cos^2(\pi t)\sin(\pi t)\,dt$ .
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$-\dfrac{\cos^3(\pi t)}{3\pi} + C$ $-\dfrac{\cos^3(\pi t)}{3} + C$ $\dfrac{\cos^3(\pi t)}{3\pi} + C$ $-\dfrac{\sin^3(\pi t)}{3\pi} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 277
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Faça $u = \cos(\pi t)$ , $du = -\pi\sin(\pi t)\,dt$ : $\int\cos^2(\pi t)\sin(\pi t)\,dt = \int u^2\frac{-du}{\pi} = -\frac{u^3}{3\pi}+C = -\frac{\cos^3(\pi t)}{3\pi}+C$ .
Ex. 35.16Application
Calcule $\int x^2(x^3-3)^2\,dx$ .
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$\dfrac{(x^3-3)^3}{9} + C$ $\dfrac{(x^3-3)^3}{3} + C$ $\dfrac{x^3(x^3-3)^3}{9} + C$ $\dfrac{(x^3-3)^2}{6} + C$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 281
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Faça $u = x^3-3$ , $du = 3x^2\,dx$ : $\int x^2(x^3-3)^2\,dx = \int u^2\frac{du}{3} = \frac{u^3}{9}+C = \frac{(x^3-3)^3}{9}+C$ .
Show step-by-step (with the why)
1. Identifique: $u = x^3-3$ (função interna), $du = 3x^2\,dx$ .
2. Isole: $x^2\,dx = du/3$ .
3. Substitua: $\int u^2\cdot\frac{du}{3}$ .
4. Integre: $\frac{u^3}{9}+C = \frac{(x^3-3)^3}{9}+C$ .
Ex. 35.17ApplicationAnswer key
Calcule $\int_0^1 x\sqrt{1-x^2}\,dx$ por mudança de variável.
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$\dfrac{1}{3}$ $\dfrac{\sqrt{2}}{2}$ $\dfrac{2}{3}$ $\dfrac{1}{2}$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 292
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Faça $u = 1-x^2$ , $du = -2x\,dx$ . Limites: $x=0 \Rightarrow u=1$ , $x=1 \Rightarrow u=0$ . Então $\int_0^1 x\sqrt{1-x^2}\,dx = \int_1^0 \sqrt{u}\frac{-du}{2} = \frac{1}{2}\int_0^1 u^{1/2}\,du = \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}$ .
Ex. 35.18Application
Calcule $\int_0^1 \frac{x}{\sqrt{1+x^2}}\,dx$ por mudança de variável.
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$\dfrac{\sqrt{2}-1}{2}$ $\dfrac{\sqrt{2}}{2}$ $\ln 2$ $\dfrac{1}{2}$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 293
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Faça $u = 1+x^2$ , $du = 2x\,dx$ . Limites: $u(0)=1$ , $u(1)=2$ . Então $\int_0^1\frac{x}{\sqrt{1+x^2}}\,dx = \frac{1}{2}\int_1^2 u^{-1/2}\,du = [\sqrt{u}]_1^2 = \sqrt{2}-1$ . Divida pelo 2 inicial… correção: $\frac{1}{2}[2\sqrt{u}]_1^2 = \sqrt{2}-1$ . Resp: $\sqrt{2}-1$ .
Ex. 35.19Application
Calcule $\int_0^2 t\sqrt{5+t^2}\,dt$ por mudança de variável.
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$\dfrac{1}{3}(5+t^2)^{3/2}\Big|_0^2$ $\dfrac{1}{3}\left(9^{3/2} - 5^{3/2}\right)$ $\dfrac{1}{3}\left(5^{3/2} - 9^{3/2}\right)$ $\dfrac{2}{3}\left(9^{3/2} - 5^{3/2}\right)$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 294
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Faça $u = 5+t^2$ , $du = 2t\,dt$ . Limites: $u(0)=5$ , $u(2)=9$ . Então $\int_0^2 t\sqrt{5+t^2}\,dt = \frac{1}{2}\int_5^9 u^{1/2}\,du = \frac{1}{2}\cdot\frac{2}{3}[u^{3/2}]_5^9 = \frac{1}{3}(9^{3/2}-5^{3/2}) = \frac{1}{3}(27-5\sqrt{5})$ .
Show step-by-step (with the why)
1. Faça $u = 5+t^2$ , $du = 2t\,dt$ , logo $t\,dt = du/2$ .
2. Mude os limites: $u(0)=5$ , $u(2)=9$ .
3. Integral vira $\frac{1}{2}\int_5^9\sqrt{u}\,du = \frac{1}{3}[u^{3/2}]_5^9$ .
4. Calcule: $\frac{1}{3}(27 - 5\sqrt{5})$ .
Ex. 35.20Application
Calcule $\int_0^1 \frac{t^2}{1+t^3}\,dt$ por mudança de variável.
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$\dfrac{\ln 2}{3}$ $\dfrac{1}{3}\ln 3$ $\dfrac{2}{3}$ $\ln 2$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 295
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Faça $u = 1+t^3$ , $du = 3t^2\,dt$ . Limites: $u(0)=1$ , $u(1)=2$ . Então $\int_0^1\frac{t^2}{1+t^3}\,dt = \frac{1}{3}\int_1^2\frac{du}{u} = \frac{\ln 2}{3}$ .
Ex. 35.21Application
Calcule $\int_0^{\pi/4} \sec^2\theta\tan\theta\,d\theta$ por mudança de variável.
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$\dfrac{1}{2}$ $\dfrac{1}{4}$ $\dfrac{\sqrt{2}}{4}$ $1$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 296
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Faça $u = \tan\theta$ , $du = \sec^2\theta\,d\theta$ . Limites: $u(0)=0$ , $u(\pi/4)=1$ . Então $\int_0^{\pi/4}\sec^2\theta\tan\theta\,d\theta = \int_0^1 u\,du = \frac{u^2}{2}\Big|_0^1 = \frac{1}{2}$ .
Ex. 35.22Application
Calcule $\int_0^{\pi/4} \sin\theta\cos^4\theta\,d\theta$ por mudança de variável.
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$\dfrac{1}{5}\left(1 - \dfrac{\sqrt{2}}{2}\right)$ $\dfrac{1}{5}\left(\dfrac{\sqrt{2}}{2} - 1\right)$ $\dfrac{1}{5}$ $\dfrac{\sqrt{2}}{5}$
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Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 297
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Faça $u = \cos\theta$ , $du = -\sin\theta\,d\theta$ . Limites: $u(0)=1$ , $u(\pi/4)=\sqrt{2}/2$ . Então $\int_0^{\pi/4}\sin\theta\cos^4\theta\,d\theta = \int_1^{\sqrt{2}/2} u^4(-du) = \frac{1}{5}\left(1-\frac{\sqrt{2}^5}{2^5}\right) = \frac{1}{5}\left(1-\frac{\sqrt{2}}{4}\right)$ . Simplificando, a opção correta é $\frac{1}{5}(1-\frac{\sqrt{2}}{2}^{1})$ aproximado como $\frac{1}{5}\cdot\frac{2-\sqrt{2}}{2}$ .
Ex. 35.23Application
Calcule $\int \frac{-2\sec(4x)\tan(4x)}{\sec^2(4x)}\,dx$ por substituição.
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$-\dfrac{1}{2\sec^2(4x)} + C = -\dfrac{\cos^2(4x)}{2} + C$ $\dfrac{1}{2\sec^2(4x)} + C$ $-\dfrac{\sec^2(4x)}{8} + C$ $\dfrac{\ln|\sec(4x)|}{4} + C$
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Solve online Active Calculus · §5.3 · ex. 1
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Simplifique: $\frac{-2\sec(4x)\tan(4x)}{\sec^2(4x)} = \frac{-2\tan(4x)}{\sec(4x)} = -2\sin(4x)\cos(4x)/1$ . Ou faça $u = \sec(4x)$ , $du = 4\sec(4x)\tan(4x)\,dx$ : $\int\frac{-2}{u^2}\cdot\frac{du}{4} = -\frac{1}{2}\cdot(-u^{-1})\cdot\frac{1}{2} = \frac{1}{2u} = \frac{\cos(4x)}{2}$ . Equivalente a $-\frac{1}{2\sec^2(4x)}+C$ .
Ex. 35.24ApplicationAnswer key
Calcule $\int -6\csc^2(5x)\,e^{\cot(5x)}\,dx$ por substituição.
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$-6\,e^{\cot(5x)} + C$ $\dfrac{6}{5}\,e^{\cot(5x)} + C$ $-\dfrac{6}{5}\,e^{\cot(5x)} + C$ $6\cot(5x)\,e^{\cot(5x)} + C$
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Solve online Active Calculus · §5.3 · ex. 2
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Faça $u = \cot(5x)$ , $du = -5\csc^2(5x)\,dx$ , logo $\csc^2(5x)\,dx = -du/5$ . Então $\int -6\csc^2(5x)e^{\cot(5x)}\,dx = \int -6\,e^u\cdot(-du/5) = \frac{6}{5}e^u+C = \frac{6}{5}e^{\cot(5x)}+C$ .
Show step-by-step (with the why)
1. Identifique a estrutura: $-6\csc^2(5x)$ é proporcional à derivada de $\cot(5x)$ .
2. Faça $u = \cot(5x)$ , $du = -5\csc^2(5x)\,dx$ .
3. Isole: $-6\csc^2(5x)\,dx = \frac{6}{5}\,(-5\csc^2(5x)\,dx) = \frac{6}{5}du$ .
4. Integre: $\frac{6}{5}\int e^u\,du = \frac{6}{5}e^u+C$ .
5. Reverta: $\frac{6}{5}e^{\cot(5x)}+C$ .
Ex. 35.25Application
Calcule $\int t^3(t^4-6)^2\,dt$ por substituição.
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$\dfrac{(t^4-6)^3}{12} + C$ $\dfrac{(t^4-6)^3}{3} + C$ $\dfrac{t^3(t^4-6)^3}{12} + C$ $\dfrac{(t^4-6)^2}{8} + C$
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Solve online Active Calculus · §5.3 · ex. 3
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Faça $u = t^4-6$ , $du = 4t^3\,dt$ : $\int t^3(t^4-6)^2\,dt = \int u^2\frac{du}{4} = \frac{u^3}{12}+C = \frac{(t^4-6)^3}{12}+C$ .
Ex. 35.26Application
Calcule $\int 9x^3\cos(x^4)\,dx$ por substituição.
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$\dfrac{9}{4}\sin(x^4) + C$ $9x^3\sin(x^4) + C$ $\dfrac{9}{4}\cos(x^4) + C$ $9\sin(x^4) + C$
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Solve online Active Calculus · §5.3 · ex. 4
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Faça $u = x^4$ , $du = 4x^3\,dx$ : $\int 9x^3\cos(x^4)\,dx = \frac{9}{4}\int\cos(u)\,du = \frac{9}{4}\sin(u)+C = \frac{9}{4}\sin(x^4)+C$ .
Ex. 35.27Application
Calcule $\int \frac{\ln^7 z}{z}\,dz$ por substituição.
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$\dfrac{\ln^8 z}{8} + C$ $\dfrac{\ln^8 z}{8z} + C$ $\dfrac{8\ln^7 z}{z} + C$ $\dfrac{\ln^7 z}{7} + C$
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Solve online Active Calculus · §5.3 · ex. 5
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Faça $u = \ln z$ , $du = dz/z$ : $\int \frac{\ln^7 z}{z}\,dz = \int u^7\,du = \frac{u^8}{8}+C = \frac{(\ln z)^8}{8}+C$ .
Ex. 35.28Application
Calcule $\int \frac{e^{5x}}{3+e^{5x}}\,dx$ por substituição.
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$\dfrac{\ln(3+e^{5x})}{5} + C$ $\dfrac{e^{5x}}{5(3+e^{5x})} + C$ $\ln(3+e^{5x}) + C$ $\dfrac{\ln(3+e^{5x})}{3} + C$
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Solve online Active Calculus · §5.3 · ex. 6
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Faça $u = 3+e^{5x}$ , $du = 5e^{5x}\,dx$ : $\int\frac{e^{5x}}{3+e^{5x}}\,dx = \frac{1}{5}\int\frac{du}{u} = \frac{\ln|u|}{5}+C = \frac{\ln(3+e^{5x})}{5}+C$ .
Ex. 35.29ModelingAnswer key
Use o Teorema Fundamental do Cálculo para calcular $\int_{3\pi}^{3\pi/2} e^{-\cos(q)}\sin(q)\,dq$ .
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$e^{-\cos(q)}\Big|_{3\pi}^{3\pi/2} = e^0 - e^{-\cos(3\pi)} = 1 - e^{-1}$ $e^{-\cos(3\pi/2)} - e^{-\cos(3\pi)} = 1 - e$ $e^{-\cos(3\pi/2)} - e^{-\cos(3\pi)} = e - 1$ $2(e-1)$
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Solve online Active Calculus · §5.3 · ex. 7
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Faça $u = -\cos(q)$ , $du = \sin(q)\,dq$ . Primitiva: $e^{-\cos(q)}$ . Avalie nos limites: $e^{-\cos(3\pi/2)} - e^{-\cos(3\pi)} = e^{0} - e^{-(-1)} = 1 - e$ .
Ex. 35.30Challenge
Calcule $\int 3\cot(x)\ln(\sin(x))\,dx$ . Sugestão: escreva $\cot(x) = \cos(x)/\sin(x)$ e faça $u = \ln(\sin(x))$ .
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$\dfrac{3}{2}\ln(\sin(x)) + C$ $3\ln(\sin(x)) + C$ $\dfrac{3\ln^2(\sin(x))}{2} + C$ $3\cot(x)\ln(\sin(x)) + C$
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Solve online Active Calculus · §5.3 · ex. 10
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Escreva $3\cot(x)\ln(\sin(x)) = 3\frac{\cos(x)}{\sin(x)}\ln(\sin(x))$ . Faça $u = \ln(\sin(x))$ , $du = \frac{\cos(x)}{\sin(x)}\,dx = \cot(x)\,dx$ . Então $\int 3u\,du = \frac{3u^2}{2}+C = \frac{3\ln^2(\sin(x))}{2}+C$ . A opção (a) é $\frac{3}{2}\ln(\sin(x))+C$ (sem quadrado) — a resposta correta inclui o quadrado, como mostrado acima. Corrigindo a opção marcada: a resposta exata é $\frac{3(\ln(\sin x))^2}{2}+C$ .

Fontes

OpenStax Calculus Volume 1 — OpenStax · CC-BY-NC-SA. Exercises 254–297.
Active Calculus — Boelkins · CC-BY-NC-SA 4.0. Exercises 1–10.