Question

cómo resolver este tipo de integrales?

Answer 1

           $\displaystyle I=\int \dfrac{\cos x}{\sqrt[3]{\sin^7\,x\cos^8 x}}\; dx\\ \\ \\ \text{Cambio de variable}\\ \\ t=\sin x\to dt = \cos x\; dx\\ \\ \cos^8x=(1-\sin^2x)^4=(1-t^2)^4\\ \\ \\ I=\int \dfrac{dt}{\sqrt[3]{t^7(1-t^2)^4}}\\ \\ \\ \text{Sea }t=\sqrt{u}\to dt = \dfrac{du}{2\sqrt{u}}\\ \\ \\ I=\dfrac{1}{2}\int u^{-5/3}(1-u)^{-4/3}\,du\\ \\ \text{Sea: }z=\left(\dfrac{1-u}{u}\right)^{1/3}\\ \\ \text{Despejando y diferenciando...}\\ \\ du=-\dfrac{3z^2}{(z^3+1)^2}\,dz\\ \\ \text{Entonces:}$
 
                $\displaystyle I=\dfrac{1}{2}\int \dfrac{z^3+1}{z^2}\,dz\\ \\ \\ I=\dfrac{1}{2}\left(\dfrac{z^2}{2}-\dfrac{1}{z}\right)\\ \\ \\ \text{Por ende}\\ \\ \boxed{I=-\dfrac{3}{4}\sqrt[3]{\cot^4x}+\dfrac{3}{2}\sqrt[3]{\tan^2x}+C}$