Question

\[\int_{}^{} \sin^3 2x~~dx\]
so uh
how do i do this?

Answer 1

\[\int_{}^{} (sin^2 2x) * (sin 2x) dx \]
\[\int_{}^{} 1-cos^2 2x * \sin 2x ~dx \]
i got this far..

Answer 2

use
\[\sin3x=3sinx-4\sin^3x\\sin^3x=\frac{ 3sinx-\sin3x }{ 4 }\]
well known identity

Answer 3

so \[\sin^3(2x)=\frac{ 3\sin(2x)-\sin(6x) }{ 4 }\]

Answer 4

oh.

Answer 5

\[\int\limits_{}\sin^3(2x)=\frac{ 3 }{ 4 }\int\limits_{}\sin(2x)dx-\frac{ 1 }{ 4 }\int\limits_{}\sin(6x)dx\]
easy now :)

Answer 6

yeah got it from here,
tnx

Answer 7

welcome :)