Question

alguien me puede ayudar con la siguiente integral ?

$\int\limits^._ . {sen^{4}x.cos^{2}x } \, dx$

Answer 1

$\int \sin ^4\left(x\right)\cos ^2\left(x\right)dx$
Por reducción de integrales:
$=-\frac{\sin ^3\left(x\right)\cos ^3\left(x\right)}{6}+\frac{3}{6}\cdot \int \:\sin ^2\left(x\right)\cos ^2\left(x\right)dx$
Simplificando, usando identidad y suma:
$=\frac{1}{8}\left(\int \:1dx-\int \cos \left(4x\right)dx\right)$
Sacamos la constante, integración y sustitución:
$=-\frac{\sin ^3\left(x\right)\cos ^3\left(x\right)}{6}+\frac{3}{6}\cdot \frac{1} {8}\left(x-\frac{1}{4}\sin \left(4x\right)\right)$
Se simplifica y se añade una constante:
$=-\frac{1}{6}\sin ^3\left(x\right)\cos ^3\left(x\right)+\frac{1}{16}\left(x-\frac{1}{4}\sin \left(4x\right)\right) + C$

Answer 2

$\displaystyle I= \int {\sin^{4}x\cos^{2}x }~ dx \\ \\ I= \int {\sin^{2}x(\sin x\cos x )^2 }~ dx \\ \\ \\ I= \dfrac{1}{4}\int {\sin^{2}x(2\sin x\cos x )^2 }~ dx \\ \\ \\ I= \dfrac{1}{4}\int {\sin^{2}x\sin^22 x }~ dx \\ \\ \\ I=\dfrac{1}{4}\int \left(\dfrac{1-\cos 2x}{2}\right)\left(\dfrac{1-\cos 4x}{2}\right)~dx\\ \\ \\ I=\dfrac{1}{16}\int 1-\cos 2x-\cos 4x+\cos2x\cos4x~dx$


$\displaystyle I=\dfrac{1}{16}\left(x-\int \cos 2x~dx-\int\cos 4x~dx+\int\cos2x\cos4x~dx\right)\\ \\ \\ I=\dfrac{1}{16}\left(x-\dfrac{1}{2}\sin 2x-\dfrac{1}{4}\sin 4x+\int\dfrac{1}{2}\cos 2x+\dfrac{1}{2}\cos6x~dx\right)\\ \\ \\ I=\dfrac{x}{16}-\dfrac{1}{32}\sin 2x-\dfrac{1}{64}\sin 4x+\dfrac{1}{32}\int \cos 2x+\dfrac{1}{32}\int \cos 6x$


$I=\dfrac{x}{16}-\dfrac{1}{32}\sin 2x-\dfrac{1}{64}\sin 4x+\dfrac{1}{64}\sin 2x+\dfrac{1}{192}\sin 6x+C\\ \\ \\ \boxed{I=\dfrac{x}{16}-\dfrac{1}{32}\sin 2x-\dfrac{1}{64}\sin 4x+\dfrac{1}{192}\sin 6x+C}$