Question

derive using FTC,
the integral of sinx to cosx of (1+v^2)^10

Answer 1

\[\huge \int\limits_{\sin x}^{\cos x} (1+v^2)^{10} dv =\]

Answer 2

yes

Answer 3

Not sure how the FTC is going to help here... u-substitution would.

Answer 4

Nm, u-sub won't.

Answer 5

FTC is the fact that we're using antiderivatives, @agent0smith ? :D

Answer 6

Yeah, pretty much. Usually these type of questions, it relates to a method that makes it simple. All it tell us here is "F you, find a way to integrate this!"

Answer 7

So you would need to find f[h(x)]h'(x) - f[g(x)]g'(x)
So f[h(x)] would be (1+cos^2(x))^10
h'(x) = -sinx
f[g(x)] = (1+sin^2(x))^10
g'(x) = cosx
so just following the formula, I would have:
-sinx(1+cos^2(x))^10 - cosx(1+sin^2(x))^10
Unless I'm missing something, this would be the method of FTC.

Answer 8

thats what i got but I'm not sure if i was right

Answer 9

Yeah I don't think that's correct, unless the original question was \[\huge \frac{ d }{ dx } \int\limits\limits_{\sin x}^{\cos x} (1+v^2)^{10} dv =\]
hence my first post confirming what it was...

Answer 10

then..? how do you work it out, question is worded "use part 1 of dftc to find the derivative of the problem

Answer 11

Yeah, that was what I assumed the equation was. Either way, the method I used would have been part 1 of the FTC you mentioned.

Answer 12

You did not put the word "derivative" in your original question :P

Here's the answer to your original question. @terenzreignz it looks fun don't it? http://www.wolframalpha.com/input/?i=%5Cint%5Climits_%7B%5Csin+x%7D%5E%7B%5Ccos+x%7D+%281%2Bv%5E2%29%5E%7B10%7D+dv+%3D

Answer 13

derive using FTC,
the derivative of the integral of sinx to cosx of (1+v^2)^10

The word derivative makes a big difference here.

Answer 14

.......awesome....., i apologize for my typo

Answer 15

Btw it looks like if you did want to just integrate the original problem, you'd first have to expand out the (1+v^2)^10, using the binomial theorem, then integrate.

Answer 16

because of the ftc part, i didn't jump in to doing that, but yes

Answer 17

I thought maybe a trig substitution would help, but i don't think so. But with the derivative part added in, it's thankfully much easier and quicker.