Question

Little Review: what is the integral of sin^3(2x)

Answer 1

all right, \[\sin^3(2x)\]

Answer 2

seems like i need to do substitution rule, then use trig identities?

Answer 3

right, let \[u=2x\], so \[du = 2 dx\], so \[dx=(1/2)du\]

Answer 4

\[\int\limits_{}^{}\sin^3 x\]

Answer 5

forgot the dx ;), so that is equal to \[(1/2)\int\limits_{}^{}\sin^3(u)du\]

Answer 6

\[sin^2x=1-cos^2x\], so we'll have \[(1/2)\int\limits_{}^{}\sin^3(u)du\] = \[(1/2)\int\limits_{}^{}\sin^2(u)*sin(u)du\] = \[(1/2)\int\limits_{}^{}(1-sin^2(u))sin(u)du\]

Answer 7

so we can apply substitution rule again for \[v=\sin(u)\]

Answer 8

\[dv = \cos(u)du\], so \[(1/2)\int\limits_{}^{}(1-v^2)dv\]

Answer 9

wait, shouldn't it be 1-cos^2(u) above?

Answer 10

ahhh, yesss... sorry: \[(1/2)\int\limits_{}^{}(1-cos^2(u))sin(u)du\]

Answer 11

so v=cos(u), which is dv=-sin(u)du

Answer 12

i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du = (-1)*(1/2) \int\limits_{}^{} (1-v^2)dv = (-1)(1/2) [ v-(1/3)v^3 + C]\]

Answer 13

i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du\]
=
\[(-1)*(1/2) \int\limits_{}^{} (1-v^2)dv\]
=
\[(-1)(1/2) [ v-(1/3)v^3 + C]\]

Answer 14

so plug back in and get v = cos u --> cos (2x),
\[(-1)(1/2) [ v-(1/3)v^3 + C] = (-1)(1/2) [ \cos(2x) - (1/3)\cos^3(2x) + C]\]
\[ = (-1)(1/2) [ cos(2x) - (1/3)cos^3(2x) + C]\]