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6.2: Integrales trigonométricas
https://espanol.libretexts.org/Matematicas/Libro%3A_Calculo_elemental_(Corral)/06%3A_M%C3%A9todos_de_Integraci%C3%B3n/6.02%3A_Integrales_trigonom%C3%A9tricas
Solución: Reemplazar $\cos^2 x$ por $1 - \sin^2 x$ , luego dejar que $u=\sin\,x$ así $\du=\cos\,x\;\dx$ : \[\begin{aligned} \int \sin^2 x\;\cos^3 x~~\dx ~&=~ \int \sin^2 x ~(\cos^2 x) \;\cos\,x~\dx\...Solución: Reemplazar $\cos^2 x$ por $1 - \sin^2 x$ , luego dejar que $u=\sin\,x$ así $\du=\cos\,x\;\dx$ : \[\begin{aligned} \int \sin^2 x\;\cos^3 x~~\dx ~&=~ \int \sin^2 x ~(\cos^2 x) \;\cos\,x~\dx\ \ [6pt] &=~\ int\ underbracket [0.3pt] {\ sin^2 x ~ (1 -\ sin^2 x)\ vphantom {\ frac {1} {2}}} _ {p (u)}\;\ underbracket [0.3pt] {\ cos\, x~\ dx\ vphantom {\ frac {1} {2}} _ {\ du}\ \ [3pt] &=~\ int (u^2 - u^4)\;\ du\ \ [6pt] &=~\ frac {1} {3}\, u^3 ~-~\ frac {1} {5}\, u^5 ~+~ C\ \ [4pt] &=~\ frac…

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