anonymous
  • anonymous
Little Review: what is the integral of sin^3(2x)
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
all right, \[\sin^3(2x)\]
anonymous
  • anonymous
seems like i need to do substitution rule, then use trig identities?
anonymous
  • anonymous
right, let \[u=2x\], so \[du = 2 dx\], so \[dx=(1/2)du\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[\int\limits_{}^{}\sin^3 x\]
anonymous
  • anonymous
forgot the dx ;), so that is equal to \[(1/2)\int\limits_{}^{}\sin^3(u)du\]
anonymous
  • anonymous
\[sin^2x=1-cos^2x\], so we'll have \[(1/2)\int\limits_{}^{}\sin^3(u)du\] = \[(1/2)\int\limits_{}^{}\sin^2(u)*sin(u)du\] = \[(1/2)\int\limits_{}^{}(1-sin^2(u))sin(u)du\]
anonymous
  • anonymous
so we can apply substitution rule again for \[v=\sin(u)\]
anonymous
  • anonymous
\[dv = \cos(u)du\], so \[(1/2)\int\limits_{}^{}(1-v^2)dv\]
anonymous
  • anonymous
wait, shouldn't it be 1-cos^2(u) above?
anonymous
  • anonymous
ahhh, yesss... sorry: \[(1/2)\int\limits_{}^{}(1-cos^2(u))sin(u)du\]
anonymous
  • anonymous
so v=cos(u), which is dv=-sin(u)du
anonymous
  • anonymous
i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du = (-1)*(1/2) \int\limits_{}^{} (1-v^2)dv = (-1)(1/2) [ v-(1/3)v^3 + C]\]
anonymous
  • anonymous
i think I got it now: so\[(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du\] = \[(-1)*(1/2) \int\limits_{}^{} (1-v^2)dv\] = \[(-1)(1/2) [ v-(1/3)v^3 + C]\]
anonymous
  • anonymous
so plug back in and get v = cos u --> cos (2x), \[(-1)(1/2) [ v-(1/3)v^3 + C] = (-1)(1/2) [ \cos(2x) - (1/3)\cos^3(2x) + C]\] \[ = (-1)(1/2) [ cos(2x) - (1/3)cos^3(2x) + C]\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.