Question

Por favor me podrian ayudar con esto:
Resolver las siguientes integrales enunciando claramente la técnica o propiedad usada.

Answer 1

1) Note que $d(e^x)=e^xdx$

$\displaystyle I=\int (6x^2e^x)dx\\ \\ \\ I=\int (6x^2)d(e^x)\\ \\ \\ \texttt{Recuerde ...} \int u~dv=uv-\int v~du\\ \\ I=6x^2e^x-\int e^xd(6x^2)\\ \\ I=6x^2e^x-\int e^x\cdot 12x~dx\\ \\ I=6x^2e^x-12\int x~d(e^x)\\ \\ \\ I=6x^2e^x-12\left[xe^x-\int e^x ~dx\right]\\ \\ \\ I=6x^2e^x-12(xe^x-e^x)+C\\ \\ I=6x^2e^x-12xe^x+12e^x+C\\ \\ \boxed{I=(6x^2-12x+12)e^x+C\\ \\}$


2)

$\displaystyle I=\int \cos^4 x\sin^3x~dx\\ \\ \\ \texttt{Note que }d(\cos x)=-\sin x ~dx\\ \\ \\ I=-\int \cos^4 x\sin^2x\cdot(-\sin x)~dx\\ \\ \\ I=-\int \cos^4 x\sin^2x~d(\cos x)\\ \\ \\ I=-\int \cos^4 x(1-\cos ^2x)~d(\cos x)\\ \\ \\ \texttt{Haciendo }u=\cos x\\ \\ \\ I=-\int u^4(1-u^2)~du\\ \\ \\ I=-\int u^4-u^6~du\\ \\ \\ I=-\dfrac{u^5}{5}+\dfrac{u^7}{7}+C$

$\displaystyle I=-\dfrac{\cos^5x}{5}+\dfrac{\cos^7x}{7}+C\\ \\ \\ \texttt{Luego:}\\ \\ \int\limits_{0}^{\pi/2}\cos^4 x\sin^3x~dx=\left.\left(-\dfrac{\cos^5x}{5}+\dfrac{\cos^7x}{7}\right)\right|_{0}^{\pi/2}\\ \\ \\ \boxed{\int\limits_{0}^{\pi/2}\cos^4 x\sin^3x~dx=\dfrac{2}{35}}$