Trig Integral Practice | MIT 18.01SC Single Variable Calculus, Fall 2010

[AZNsensation2424]

cool vid! I don't understand on the 2nd problem how you got ~sinx(2cos(^2)x-1)dx = to -(2/3)cos(^3)x + cosx + C. What I don't understand is what happened to the sinx, it looks like you distribute it but i don't see how 2sinxcos(^2)x would integrate to -(2/3)cos(^3)x. If you or someone else could explain, that would be great!

[leopard00wolf]

@AZNsensation2424 It's just the chain rule in reverse: when you take a derivative of cos^3(x), you get 3 * cos^2(x) * (-sin(x)), where the last factor comes from the chain rule. (Equivalently, make the substitution u = cos(x), du = -sin(x) dx.)

[Jnorrisclay]

wait....this is awesome...how does this have like no views