∫ cos(sinx).dx=?
2007-03-14 01:07:58 · 6 answers · asked by mockingbird 1 in Science & Mathematics ➔ Mathematics
may be it can be solved if it is a question of definite integral with some limit of integration, say from 0 to pi/2. But in this form it appears to be an impossible integral. Pl don't waste your time trying to solve it.
2007-03-14 07:00:17 · answer #1 · answered by makeitsimple 2 · 0⤊ 0⤋
Well, let's get rid of sin x:
Let u = sin x, x = arcsin u, dx = dx/â(1-x²)
So now we have to integrate
cos u du/ â(1-u²).
But here's where our luck runs out. This
last integral is not elementary. In fact,
the Wolfram integrator could not find a formula for it.
2007-03-14 15:39:23 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
Are you sure it is of the form â«cos[sin(x)]dx and not â«cos(x)*sin(x)dx ?
Anyway, here's a link that may give an answer, but not how to get it.
http://integrals.wolfram.com/index.jsp
2007-03-14 09:31:27 · answer #3 · answered by RememberingCalculus 1 · 0⤊ 0⤋
Check the question. In the form you've given it I think it's impossible. Is there an "exponent" -1 somewhere? (Not really an exponent, but meaning inverse function)
2007-03-14 08:30:19 · answer #4 · answered by Hy 7 · 1⤊ 0⤋
This is a very difficult integral. If you are new to integration then I don't think that you would have been asked to attempt this one.
2007-03-14 08:33:15 · answer #5 · answered by mathsmanretired 7 · 1⤊ 0⤋
hmm... trigonometric derivatives are...
sinx = cosx
cosx = -sinx
tanx = sec^2 x
cotx = -csc^2 x
secx = secx tanx
cscx = - cscx cotx
maybe you're just supposed to look for the opposite when looking for the integral..something? i'm not sure. then add a +c at the end? hmm..........
2007-03-14 08:21:04 · answer #6 · answered by ieatreese88 2 · 0⤊ 3⤋