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

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

OCW Scholar - Single Variable Calculus
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

all right, \[\sin^3(2x)\]
seems like i need to do substitution rule, then use trig identities?
right, let \[u=2x\], so \[du = 2 dx\], so \[dx=(1/2)du\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question