Solving Integral: sin^3x cos^2x dx

[Ravenatic20]

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]

You could write everything 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]

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]

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]

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: $$+ C$$.

 

[d_leet]

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]

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]

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]

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

 

[jdavel]

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]

Err... That did not make much sense, sorry.

 

[d_leet]

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]

tx = -sin(x)?

 

[Vid]

Well no wonder you couldn't get this integral, you don't understand the most basic method of integration.

 

[Ravenatic20]

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.

 

[awvvu]

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]

$$...=-\int(1-cos^2(x))cos^2(x)d(cos(x))=-\int(1-t^2)t^2dt$$

can you do it now??

ciao
marco

 

[HallsofIvy]

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]<br /> Now, if u= cos(x), what is du? If you don't know that you should review differentiation before trying integration.[/tex]