Regra da Substituição nas Integrais
Guia prático com muitos exercícios resolvidos — passos alinhados e integrais definidas incluídas.
📘 O que é a Regra da Substituição
A regra da substituição (ou mudança de variável) transforma integrais difíceis em expressões mais simples. Fazemos \(u=g(x)\), trocamos \(dx\) por \(du\) via \(du=g'(x)\,dx\) e resolvemos em função de \(u\).
📐 Regra geral
Se \(u=g(x)\) e \(du=g'(x)\,dx\), então:
Em integrais definidas, mude também os limites: se \(x=a\Rightarrow u=g(a)\) e \(x=b\Rightarrow u=g(b)\).
🔹 Passos para aplicar
- Escolha \(u=g(x)\) (a “parte interna” ou expressão composta).
- Calcule \(du=g'(x)\,dx\) e ajuste constantes (multiplique/divida).
- Substitua completamente \(x\) e \(dx\) por \(u\) e \(du\).
- Integre em \(u\) e volte para \(x\).
- (Definidas) Converta os limites para \(u\) e não precisa voltar para \(x\).
🧮 Exemplos resolvidos (aquecimento)
Exemplo A
\(\displaystyle \int 2x\,(x^2+3)^4\,dx\)
Seja \(u=x^2+3\Rightarrow du=2x\,dx\).
\(\int 2x(x^2+3)^4\,dx=\int u^4\,du\)
\(=\dfrac{u^5}{5}+C\)
\(=\boxed{\dfrac{(x^2+3)^5}{5}+C}\)
Exemplo B
\(\displaystyle \int x^3(x^4+5)^{12}\,dx\)
\(u=x^4+5\Rightarrow du=4x^3\,dx\Rightarrow x^3\,dx=\dfrac{du}{4}\)
\(\int x^3(x^4+5)^{12}dx=\dfrac14\int u^{12}du\)
\(=\dfrac{u^{13}}{52}+C\)
\(=\boxed{\dfrac{(x^4+5)^{13}}{52}+C}\)
Exemplo C
\(\displaystyle \int e^{5x-2}\,dx\)
\(u=5x-2\Rightarrow du=5\,dx\Rightarrow dx=\dfrac{du}{5}\)
\(\int e^{5x-2}dx=\dfrac15\int e^u\,du\)
\(=\dfrac{e^u}{5}+C\)
\(=\boxed{\dfrac{e^{5x-2}}{5}+C}\)
Exemplo D
\(\displaystyle \int \sin^3x\,\cos x\,dx\)
\(u=\sin x\Rightarrow du=\cos x\,dx\)
\(\int \sin^3x\cos x\,dx=\int u^3\,du\)
\(=\dfrac{u^4}{4}+C\)
\(=\boxed{\dfrac{\sin^4 x}{4}+C}\)
📚 Exercícios Resolvidos (passo a passo)
-
\(\displaystyle \int (3x+1)^7\,dx\)
\(u=3x+1\Rightarrow du=3\,dx\Rightarrow dx=\dfrac{du}{3}\)
\(\int (3x+1)^7 dx=\dfrac13\int u^7 du\)
\(=\dfrac{u^8}{24}+C\)
\(=\boxed{\dfrac{(3x+1)^8}{24}+C}\)
-
\(\displaystyle \int \cos(4x)\,\sin^{3}(4x)\,dx\)
\(u=\sin(4x)\Rightarrow du=4\cos(4x)\,dx\Rightarrow \cos(4x)\,dx=\dfrac{du}{4}\)
\(\int \cos(4x)\sin^3(4x)dx=\dfrac14\int u^3 du\)
\(=\dfrac{u^4}{16}+C\)
\(=\boxed{\dfrac{\sin^4(4x)}{16}+C}\)
-
\(\displaystyle \int x\,\cos(x^2)\,dx\)
\(u=x^2\Rightarrow du=2x\,dx\Rightarrow x\,dx=\dfrac{du}{2}\)
\(\int x\cos(x^2)dx=\dfrac12\int \cos u\,du\)
\(=\dfrac12\sin u + C\)
\(=\boxed{\dfrac{\sin(x^2)}{2}+C}\)
-
\(\displaystyle \int \dfrac{2x}{x^2+9}\,dx\)
\(u=x^2+9\Rightarrow du=2x\,dx\)
\(\int \frac{2x}{x^2+9}dx=\int \frac{1}{u}du\)
\(=\ln|u|+C\)
\(=\boxed{\ln(x^2+9)+C}\)
-
\(\displaystyle \int \dfrac{3}{\sqrt{1-9x^2}}\,dx\)
\(u=3x\Rightarrow du=3\,dx\Rightarrow dx=\dfrac{du}{3}\)
\(\int \frac{3}{\sqrt{1-9x^2}}dx=\int \frac{1}{\sqrt{1-u^2}}du\)
\(=\arcsin(u)+C\)
\(=\boxed{\arcsin(3x)+C}\)
-
\(\displaystyle \int e^{x}\,\sqrt{1+e^{2x}}\,dx\)
\(u=1+e^{2x}\Rightarrow du=2e^{2x}\,dx=2(e^x)^2 dx\)
Mas \(e^x\,dx=\dfrac{du}{2e^x}\) não simplifica direto. Melhor: \(t=e^x\Rightarrow dt=e^x dx\).
\(\int e^x\sqrt{1+e^{2x}}dx=\int \sqrt{1+t^2}\,dt\)
\(\int \sqrt{1+t^2}dt=\dfrac{t}{2}\sqrt{1+t^2}+\dfrac12\ln\!\big|t+\sqrt{1+t^2}\big|+C\)
\(=\boxed{\dfrac{e^x}{2}\sqrt{1+e^{2x}}+\dfrac12\ln\!\big(e^x+\sqrt{1+e^{2x}}\big)+C}\)
-
\(\displaystyle \int \dfrac{x}{\sqrt{x^2+4}}\,dx\)
\(u=x^2+4\Rightarrow du=2x\,dx\Rightarrow x\,dx=\dfrac{du}{2}\)
\(\int \frac{x}{\sqrt{x^2+4}}dx=\dfrac12\int u^{-1/2}du\)
\(=\dfrac12\cdot \dfrac{u^{1/2}}{1/2}+C\)
\(=\boxed{\sqrt{x^2+4}+C}\)
-
\(\displaystyle \int \sec^{2}(3x)\,\tan^{5}(3x)\,dx\)
\(u=\tan(3x)\Rightarrow du=3\sec^2(3x)\,dx\Rightarrow \sec^2(3x)\,dx=\dfrac{du}{3}\)
\(\int \sec^2(3x)\tan^5(3x)dx=\dfrac13\int u^5 du\)
\(=\dfrac{u^6}{18}+C\)
\(=\boxed{\dfrac{\tan^6(3x)}{18}+C}\)
-
\(\displaystyle \int \dfrac{dx}{(2x+5)^3}\)
\(u=2x+5\Rightarrow du=2\,dx\Rightarrow dx=\dfrac{du}{2}\)
\(\int \frac{dx}{(2x+5)^3}=\dfrac12\int u^{-3}du\)
\(=\dfrac12\cdot \dfrac{u^{-2}}{-2}+C\)
\(=\boxed{-\dfrac{1}{4(2x+5)^2}+C}\)
-
\(\displaystyle \int \dfrac{5x+1}{(x^2+x+7)^2}\,dx\)
Completar quadrado: \(x^2+x+7=\big(x+\tfrac12\big)^2+\tfrac{27}{4}\). Defina \(u=x+\tfrac12\) e \(a^2=\tfrac{27}{4}\Rightarrow a=\tfrac{3\sqrt3}{2}\).
O numerador: \(5x+1=5u-\tfrac32\). Então:
\(I=\int \frac{5u-\tfrac32}{(u^2+a^2)^2}\,du=5\int \frac{u}{(u^2+a^2)^2}du-\tfrac32\int \frac{1}{(u^2+a^2)^2}du.\)
Fórmulas padrão: \(\int \frac{u}{(u^2+a^2)^2}du=-\frac{1}{2(u^2+a^2)}\) e \(\int \frac{1}{(u^2+a^2)^2}du=\frac{u}{2a^2(u^2+a^2)}+\frac{1}{2a^3}\arctan\!\big(\tfrac{u}{a}\big)\).
Aplicando e simplificando com \(a^2=\tfrac{27}{4}\), \(a^3=\tfrac{81\sqrt3}{8}\):
\(I=-\frac{5}{2(u^2+a^2)}-\frac{u}{9(u^2+a^2)}-\frac{2}{27\sqrt3}\arctan\!\big(\tfrac{u}{a}\big)+C.\)
Voltando para \(x\): \(u=x+\tfrac12\), \(u^2+a^2=x^2+x+7\), \(\tfrac{u}{a}=\tfrac{2x+1}{3\sqrt3}\).
\(\boxed{\displaystyle I=-\frac{x+23}{9\,(x^2+x+7)}-\frac{2}{27\sqrt3}\,\arctan\!\left(\frac{2x+1}{3\sqrt3}\right)+C}\)
-
\(\displaystyle \int_{0}^{1} 2x\,(x^2+4)\,dx\) (definida)
\(u=x^2+4\Rightarrow du=2x\,dx\)
Limites: \(x=0\Rightarrow u=4\);\; \(x=1\Rightarrow u=5\).
\(\int_{0}^{1}2x(x^2+4)dx=\int_{4}^{5} u\,du\)
\(=\left.\dfrac{u^2}{2}\right|_{4}^{5}\)
\(=\dfrac{25}{2}-\dfrac{16}{2}\)
\(=\boxed{\dfrac{9}{2}}\)
-
\(\displaystyle \int_{0}^{\pi/2}\sin(3x)\cos^5(3x)\,dx\) (definida)
\(u=\cos(3x)\Rightarrow du=-3\sin(3x)dx\Rightarrow \sin(3x)dx=-\dfrac{du}{3}\)
Limites: \(x=0\Rightarrow u=\cos 0=1\);\; \(x=\tfrac{\pi}{2}\Rightarrow u=\cos\!\tfrac{3\pi}{2}=0\).
\(\int_{0}^{\pi/2}\sin(3x)\cos^5(3x)dx=-\dfrac{1}{3}\int_{1}^{0} u^5 du\)
\(=\dfrac{1}{3}\int_{0}^{1} u^5 du\)
\(=\dfrac{1}{3}\cdot \left.\dfrac{u^6}{6}\right|_{0}^{1}\)
\(=\boxed{\dfrac{1}{18}}\)
-
\(\displaystyle \int \dfrac{dx}{x\ln x}\), \(x>1\)
\(u=\ln x\Rightarrow du=\dfrac{dx}{x}\)
\(\int \frac{dx}{x\ln x}=\int \frac{1}{u}\,du\)
\(=\boxed{\ln|\ln x|+C}\)
-
\(\displaystyle \int \dfrac{4x}{(x^2+1)^2}\,dx\)
\(u=x^2+1\Rightarrow du=2x\,dx\Rightarrow 4x\,dx=2\,du\)
\(\int \frac{4x}{(x^2+1)^2}dx=2\int u^{-2}du\)
\(=2\cdot \left(-u^{-1}\right)+C\)
\(=\boxed{-\dfrac{2}{x^2+1}+C}\)
-
\(\displaystyle \int_{1}^{e} \dfrac{dx}{x\,(1+\ln x)}\) (definida)
\(u=1+\ln x\Rightarrow du=\dfrac{dx}{x}\)
Limites: \(x=1\Rightarrow u=1+\ln 1=1\);\; \(x=e\Rightarrow u=1+\ln e=2\).
\(\int_{1}^{e}\frac{dx}{x(1+\ln x)}=\int_{1}^{2}\frac{1}{u}\,du\)
\(=\left.\ln|u|\right|_{1}^{2}\)
\(=\boxed{\ln 2}\)
📝 Exercícios Propostos (com solução)
Resolva usando substituição. Abra cada item para conferir a solução completa.
-
\(\displaystyle \int 6x\,(x^2-4)^5\,dx\)
\(u=x^2-4\Rightarrow du=2x\,dx\Rightarrow 6x\,dx=3\,du\)
\(\int 6x(x^2-4)^5dx=3\int u^5 du\)
\(=3\cdot \dfrac{u^6}{6}+C\)
\(=\boxed{\dfrac{(x^2-4)^6}{2}+C}\)
-
\(\displaystyle \int \dfrac{\cos(5x)}{\sqrt{1+\sin(5x)}}\,dx\)
\(u=\sin(5x)\Rightarrow du=5\cos(5x)dx\Rightarrow \cos(5x)dx=\dfrac{du}{5}\)
\(\int \frac{\cos(5x)}{\sqrt{1+\sin(5x)}}dx=\dfrac{1}{5}\int (1+u)^{-1/2}du\)
\(=\dfrac{1}{5}\cdot \dfrac{(1+u)^{1/2}}{1/2}+C\)
\(=\boxed{\dfrac{2}{5}\sqrt{1+\sin(5x)}+C}\)
-
\(\displaystyle \int \dfrac{e^{2x}}{1+e^{2x}}\,dx\)
\(u=1+e^{2x}\Rightarrow du=2e^{2x}dx\Rightarrow e^{2x}dx=\dfrac{du}{2}\)
\(\int \frac{e^{2x}}{1+e^{2x}}dx=\dfrac12\int \frac{1}{u}du\)
\(=\boxed{\dfrac12\ln(1+e^{2x})+C}\)
-
\(\displaystyle \int_{0}^{2}\dfrac{x}{1+x^2}\,dx\)
\(u=1+x^2\Rightarrow du=2x\,dx\Rightarrow x\,dx=\dfrac{du}{2}\)
Limites: \(x=0\Rightarrow u=1\);\; \(x=2\Rightarrow u=5\).
\(\int_{0}^{2}\frac{x}{1+x^2}dx=\dfrac12\int_{1}^{5}\frac{1}{u}du\)
\(=\dfrac12\left.\ln u\right|_{1}^{5}\)
\(=\boxed{\dfrac12\ln 5}\)
-
\(\displaystyle \int \dfrac{dx}{\sqrt{4-9x^2}}\), \(|x|<\tfrac{2}{3}\)
\(u=3x\Rightarrow du=3dx\Rightarrow dx=\dfrac{du}{3}\)
\(\int \frac{dx}{\sqrt{4-9x^2}}=\dfrac13\int \frac{1}{\sqrt{4-u^2}}du\)
Faça \(u=2\sin\theta\Rightarrow du=2\cos\theta d\theta\) (ou use forma padrão).
Resultado padrão: \(\int \frac{du}{\sqrt{a^2-u^2}}=\arcsin\!\frac{u}{a}+C\)
\(=\dfrac13\arcsin\!\left(\dfrac{u}{2}\right)+C\)
\(=\boxed{\dfrac13\arcsin\!\left(\dfrac{3x}{2}\right)+C}\)
-
\(\displaystyle \int (x+1)\,e^{x^2+2x}\,dx\)
\(u=x^2+2x\Rightarrow du=(2x+2)dx=2(x+1)dx\Rightarrow (x+1)dx=\dfrac{du}{2}\)
\(\int (x+1)e^{x^2+2x}dx=\dfrac12\int e^u du\)
\(=\boxed{\dfrac{e^{x^2+2x}}{2}+C}\)
-
\(\displaystyle \int \dfrac{\ln x}{x}\,dx\), \(x>0\)
\(u=\ln x\Rightarrow du=\dfrac{dx}{x}\)
\(\int \frac{\ln x}{x}dx=\int u\,du\)
\(=\boxed{\dfrac{(\ln x)^2}{2}+C}\)
-
\(\displaystyle \int \dfrac{dx}{(x+2)\sqrt{x+2}}\)
\(u=x+2\Rightarrow du=dx\)
\(\int \frac{dx}{(x+2)\sqrt{x+2}}=\int \frac{1}{u^{3/2}}du\)
\(=\left.\frac{u^{-1/2}}{-1/2}\right.+C\)
\(=\boxed{-\dfrac{2}{\sqrt{x+2}}+C}\)
-
\(\displaystyle \int_{0}^{\ln 3}\dfrac{e^{x}}{1+e^{x}}\,dx\) (definida)
\(u=1+e^{x}\Rightarrow du=e^{x}dx\)
Limites: \(x=0\Rightarrow u=2\);\; \(x=\ln 3\Rightarrow u=1+3=4\).
\(\int_{0}^{\ln 3}\frac{e^x}{1+e^x}dx=\int_{2}^{4}\frac{1}{u}du\)
\(=\left.\ln u\right|_{2}^{4}\)
\(=\boxed{\ln 2}\)
-
\(\displaystyle \int \dfrac{3x^2-1}{(x^3-x+5)^2}\,dx\)
\(u=x^3-x+5\Rightarrow du=(3x^2-1)dx\)
\(\int \frac{3x^2-1}{(x^3-x+5)^2}dx=\int u^{-2}du\)
\(=\boxed{-\dfrac{1}{x^3-x+5}+C}\)
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