Question

Calcular as integrais usando o método de substituição:

a) ∫cos(7∅+5)d∅

b)∫x²e׳dx

c)∫dx/3x-7

Answer 1

a) $\displaystyle I=\int \cos(7\theta+5)\,d\theta$

Vamos fazer a substituição: $x=7\theta+5\to dx=7d\theta\to d\theta=\dfrac{1}{7}dx$. Usando na integral acima:

$\displaystyle I=\int \cos(7\theta+5)\,d\theta\\\\ I=\int \cos(x)\,\dfrac{1}{7}dx\\\\ I=\dfrac{1}{7}\int \cos(x)\,dx\\\\ I=\dfrac{1}{7}(\sin(x)+C_1)\\\\ I=\dfrac{1}{7}\sin(x)+C\\\\ I=\dfrac{1}{7}\sin(7\theta+5)+C\\\\ \boxed{\int \cos(7\theta+5)\,d\theta=\dfrac{1}{7}\sin(7\theta+5)+C}$


b) $\displaystyle I=\int x^2e^{x^3}\,dx$

Vamos fazer a substituição: $y=x^3\to dy=3x^2\,dx\to dx=\dfrac{dy}{3x^2}$. Aplicando na integral:

$\displaystyle I=\int x^2e^{x^3}\,dx\\\\ I=\int x^2e^{y}\,\dfrac{dy}{3x^2}\\\\ I=\int e^{y}\,\dfrac{dy}{3}\\\\ I=\dfrac{1}{3}\int e^{y}\,dy\\\\ I=\dfrac{1}{3}[e^y+C_1]\\\\ I=\dfrac{1}{3}e^y+C\\\\ I=\dfrac{1}{3}e^{x^3}+C\\\\ \boxed{\int x^2e^{x^3}\,dx=\dfrac{1}{3}e^{x^3}+C}$


c) $\displaystyle I=\int \dfrac{dx}{3x-7}$

Vamos fazer a substituição: $y=3x-7\to dy=3dx\to dx=\dfrac{1}{3}dy$. Logo, a integral acima pode ser escrita como:

$\displaystyle I=\int \dfrac{dx}{3x-7}\\\\ I=\int \dfrac{\frac{1}{3}dy}{y}\\\\ I=\dfrac{1}{3}\int \dfrac{dy}{y}\\\\ I=\dfrac{1}{3}(\ln|y|+C_1)\\\\ I=\dfrac{1}{3}\ln|y|+C\\\\ I=\dfrac{1}{3}\ln|3x-7|+C\\\\ \boxed{\int \dfrac{dx}{3x-7}=\dfrac{1}{3}\ln|3x-7|+C}$