Question

∫_0^(4/π)▒〖sin^3⁡(2x) cos^4⁡(2x)dx〗

Agradeciendo mucho su amable respuesta.

Answer 1

$\displaystyle \int_{0}^{4/\pi}\sin^32x\cos^4 2x \;dx\\ \\ \text{Cambio de variable: }\\ \\ u=\cos 2x\to du =-2\sin 2x\;dx\\ \\ \int\sin^32x\cos^4 2x \;dx=-\dfrac{1}{2}\int \sin^22x\cos^42x (-2\sin 2x)dx\\ \\ \int\sin^32x\cos^4 2x \;dx=-\dfrac{1}{2}\int (1-\cos^22x)\cos^42x (-2\sin 2x)dx\\ \\ \int\sin^32x\cos^4 2x \;dx=-\dfrac{1}{2}\int (1-u^2)u^4 du$

$\displaystyle \int\sin^32x\cos^4 2x \;dx=-\dfrac{1}{2}\int u^4-u^6 du\\ \\ \int\sin^32x\cos^4 2x \;dx=-\dfrac{1}{2}\left(\dfrac{u^5}{5}-\dfrac{u^7}{7}\right)\\ \\ \int\sin^32x\cos^4 2x \;dx=\dfrac{u^7}{14}-\dfrac{u^5}{10}\\ \\ \int\sin^32x\cos^4 2x \;dx=\dfrac{\cos^72x}{14}-\dfrac{\cos^52x}{10}\\ \\ \\ \boxed{\int_0^{4/\pi}\sin^32x\cos^4 2x \;dx=\dfrac{\cos^7(8/\pi)}{14}-\dfrac{\cos^5(8/\pi)}{10}+\dfrac{1}{35}}$

Answer 2

$Hola~~Anlcy \\ \\ \int_{0}^{ \frac{4}{ \pi } } [sin ^{3}2x~cos ^{4} 2x ]dx \\ \\ cambiamos~de~variable~siendo: \\ \boxed{u=cos2x}~~--\ \textgreater \ su~derivada \\ \\ \frac{du}{dx} =-2sin2x~~--\ \textgreater \ queda: \\ \\ \boxed{ - \frac{du}{2} =sin2xdx} \\ \\ De~la~expresi\acute{o}n~hacemos: \\ \int_{0} ^{ \frac{4}{ \pi } } [sin\³2x~cos ^{4} 2x]dx~ \\ \\ \int_{0} ^{4/ \pi } sen ^{2}2x~cos ^{4} 2x~sen2xdx$

$\int _{0}^{ 4/ \pi } (1-cos\²2x)(cos ^{4} 2x)~sin2xdx~~~--\ \textgreater \ multiplicando \\ \\ \int _{0}^{4/ \pi } cos ^{4}2x~sin2xdx- \int _{0}^{4/ \pi } cos ^{6} 2x~sin2xdx \\ \\ Reemplazando~[sin2xdx=- \frac{du}{2} ]~y~[cos2x=u]~vea: \\ \\ \int _{0}^{4/ \pi } u^{4} (- \frac{du}{2} )-\int _{0}^{4/ \pi } u^{6}(- \frac{du}{2}) \\ \\ - \frac{1}{2} \int _{0}^{4/ \pi } u^{4} du+ \frac{1}{2} \int _{0}^{4/ \pi } u^{6} du~~--\ \textgreater \ integrando~tenemos:$

$- \frac{1}{2} \frac{ u^{5} }5} + \frac{1}{2} \frac{ u^{7} }{7} ~~ ] _{0}^{4/ \pi } \\ \\ - \frac{ u^{5} }{10} + \frac{ u^{7} }{14} ~~] _{0}^{4/ \pi } ~~--\ \textgreater \ reemplazamos[u=cos2x], tenemos: \\ \\ - \frac{ cos^{5}2x }{10} + \frac{ cos^{7}2x }{14} ~~] _{0}^{4/ \pi } ~~--\ \textgreater \ reemplazando ~en~[x=4/ \pi ~~y~~x=0] \\ \\ - \frac{ cos^{5} (2. \frac{4}{ \pi } )}{10} + \frac{ cos^{7}(2. \frac{4}{ \pi } ) }{14} -[- \frac{ cos^{5}(0) }{10} + \frac{ cos^{7}0 }{14} ]$

$- \frac{ cos^{5}( \frac{8}{ \pi } ) }{10} + \frac{cos ^{7}( \frac{8}{ \pi } ) }{14} -[ -\frac{1}{10}+ \frac{1}{14}] \\ \\ \boxed{\boxed{- \frac{cos ^{5}( \frac{8}{ \pi } ) }{10}+ \frac{cos ^{7}( \frac{8}{ \pi } ) }{14} + \frac{1}{35} }}$

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     Espero te sirva  saludos!