View Full Version : Integration Question
Ravenatic20
Feb19-08, 12:10 AM
I've been working on this problem for almost 4 days now and have made no progress. Once I think I've got it right, by professor says its wrong, and to try again. I've tried and tried. Any ideas? Here it is:
\int sin^{3}x cos^{2}x dx
scottie_000
Feb19-08, 12:44 AM
You could write everthing in the integral in terms of sines, then either:
1. Use reduction formulae for sin^n x
2. write the powers of sines in terms of sines of multiples of x
Mystic998
Feb19-08, 12:57 AM
Try to write the integrand as sin(x) * f(cos(x)) or cos(x) * f(sin(x)), where f is some algebraic function.
Marco_84
Feb19-08, 01:13 AM
I've been working on this problem for almost 4 days now and have made no progress. Once I think I've got it right, by professor says its wrong, and to try again. I've tried and tried. Any ideas? Here it is:
\int sin^{3}x cos^{2}x dx
I know this is not so formal, but it works:
using the fact that:
d(cos(x))=-sin(x)dx
you get:
\int sin^{3}(x) cos^{2}(x) dx=-\int sin^2(x)cos^2(x)d(cos(x))=-\int(1-cos^2(x))cos^2(x)d(cos(x))
and you are done.
regards
marco
Ravenatic20
Feb19-08, 01:33 AM
Thanks marco but I need the whole integral solved so the \int sign is removed. And to the point where our constant C is added on: + C.
Thanks marco but I need the whole integral solved so the \int sign is removed. And to the point where our constant C is added on: + C.
Do you honestly expect others to just do your work for you. Marco made the problem much simpler for you all it requires now is a simple, and fairly obvious substitution.
Ravenatic20
Feb19-08, 07:38 AM
Do you honestly expect others to just do your work for you. Marco made the problem much simpler for you all it requires now is a simple, and fairly obvious substitution.
No, the last part just doesn't make sense. If someone could explain it I'll take a shot at it, but I've never seen it (d(cos(x)))
Marco_84
Feb19-08, 08:12 AM
No, the last part just doesn't make sense. If someone could explain it I'll take a shot at it, but I've never seen it (d(cos(x))
do the substitution:
t=cos(x)-----> dt=d(cos(x))
you get it???
regards
marco
Ravenatic20
Feb19-08, 08:08 PM
Thanks Marco.
This is what I have so far, in continuation of what Marco helped out with:
=-\int(1-cos^2(x))cos^2(x)d(cos(x))
=-\int(\frac{1}{2}-\frac{1}{2} cos2x)(\frac{1}{2}+\frac{1}{2} cos2x)d(cos(x))
=-[{(\frac{1}{2}x-\frac{1}{4} sin2x)(\frac{1}{2}x+\frac{1}{4} sin2x)] + C
Err, is this right? If not how do I fix it? Thanks
No, that's not right.
Do what macro_84 said. let t = cos(x)
Now substitute t everywhere you see a cos(x) in the integrand that macro gave you (the one with nothing but cosines). it's staring you in the face.
Ravenatic20
Feb19-08, 10:00 PM
Err... That did not make much sense, sorry.
Err... That did not make much sense, sorry.
How did that not make sense? Are you familiar with integration by substitution? Make the substititution t=cos(x) and what happens?
Ravenatic20
Feb19-08, 11:29 PM
tx = -sin(x)?
Well no wonder you couldn't get this integral, you don't understand the most basic method of integration.
Ravenatic20
Feb20-08, 12:08 AM
Well no wonder you couldn't get this integral, you don't understand the most basic method of integration.
Sorry, I've only been doing this for a few weeks. I came here for help, nothing else.
I don't think he's familiar with that notation for substitution, the d(cos(x))
http://en.wikipedia.org/wiki/U-substitution#Examples
Marco_84
Feb20-08, 07:54 AM
...=-\int(1-cos^2(x))cos^2(x)d(cos(x))=-\int(1-t^2)t^2dt
can you do it now??
ciao
marco
HallsofIvy
Feb20-08, 11:39 AM
Then stop expecting people to do the problem for you. What has been suggested is that you rewrite the integral as
[tex]\int sin^2(x)cos^2(x) (sin(x)) dx= \int (1- cos^2(x))cos^2(x) (sin(x)dx[/itex]
Now, if u= cos(x), what is du? If you don't know that you should review differentiation before trying integration.
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