# Integration Problem

1. Dec 12, 2007

### jra_1574

1. The problem statement, all variables and given/known data

f:=sin^3(x)cos(y)
Integrate f, from x=y....pi/2)

2. Relevant equations

I can do it if the two letters were y... But i have no idea how to solve when they are combined..

Last edited: Dec 12, 2007
2. Dec 12, 2007

### HallsofIvy

Staff Emeritus
That is:
$$\int_y^{\pi/2} sin^3(x)cos^3(y) dx$$
Find the anti-derivative (there's a standard method for odd powers of sin or cos) and evaluate between $\pi/2$ and y. Of course, the answer will not be number but will depend on y.

3. Dec 12, 2007

### cristo

Staff Emeritus
Integrate f with respect to what? Is this a double integration, or a single integration? Please state the question exactly as it is given.

4. Dec 12, 2007

### jra_1574

f:=sin^3(x)*cos(y);

Integrate f with respect to x.... from y to Pi/2

Last edited: Dec 12, 2007
5. Dec 13, 2007

### Kreizhn

In all fairness to jra_1574, I wouldn't say that the question was ambiguous at all. The variable of integration may not have been explicitly stated but the way it was written

"from x=y...Pi/2"

seemed fairly obvious to me, and is in fact the way that some CAS's denote a definite integral (the primary example in mind would be Maple).

As for the actual question jra_1574, the integration doesn't depend on y, and so it's a constant and by linearity can be pulled out of the integral.

6. Dec 13, 2007

### HallsofIvy

Staff Emeritus
Agree with everything you say except for the last line! y is the lower limit of integration and cannot be "pulled out of the integral".

7. Dec 13, 2007

### Office_Shredder

Staff Emeritus
What if instead it was written as f=sin(x)^3cos(0) and was integrated from 0 to pi/2. Couldn't the cos(0) be pulled out of the integral? I'm not sure how this is supposed to be any different

8. Dec 13, 2007

### cristo

Staff Emeritus
That may be true, but the fact that he said "I don't know what to do when there are two letters" implied that it could be a double integration. Besides, one should always state explicitly what one is doing.
What you mean is that y is a constant when integrating wrt x, and so cos(y) can be treated as a constant.

9. Dec 13, 2007

### HallsofIvy

Staff Emeritus
Since cos(y) is a constant (with respect to x) it can be "pulled out of the integral". That is different from saying y (which is also the lower limit of integration) can be "pulled out of the integral".

$$\int_y^{\pi/2} sin^3(x)cos(y)dx= cos(y)\int_y^{\pi/2} sin^3(x) dx$$

10. Dec 13, 2007

### Office_Shredder

Staff Emeritus
I assumed that was what Kreizhn meant. Never mind then

11. Dec 13, 2007

### Kreizhn

Yeah, sorry, meant the cos(y). It was late and I figured it was pretty obvious what I meant

Edit: Hurrah for semantical trivialities. I know that I should've stated that better but come on HallsOfIvy, I would've imagined that you knew precisely what I was talking about and was nit-picking

Last edited: Dec 13, 2007
12. Dec 13, 2007

### HallsofIvy

Staff Emeritus
Actually, I was taken aback by it because I had focused on the "y" as a limit of the integral rather than the "cos(y)" while you were doing the opposite!

Also I agree that that looks like "half" of a double integral!

13. Dec 13, 2007

### jra_1574

How would i type that in maple?

f:=sin(x)^3*cos(y)?
or f:=sin^3(x)*cos(y)?

anyways i pulled the cos(y) out of the integral, then change sin^3(x) to sin(x)*(1-cos^2(x))... i think that led me to
sin(x)-cos^2(x)*sin(x)

so integrate sin(x)= cos(x) minus
integration of cos^2(x)*sin(x) = i am not sure but is this g and g'? or i have to choose one to be u and do the du/dx?

14. Dec 14, 2007

### Kreizhn

Don't forget that there's a $\frac{1}{2}$ in there as well.

$$sin^2(x) = \displaystyle\frac{1}{2} \left( 1-cos(2x) \right)$$

then let $u = cos(x)$ so that $du = - sin(x) dx$ and proceed from there.

In maple, you could write either

f:=sin(x)^3*cos(y);

or if you actually wanted to be able to evaluate f and certain points

f:=(x,y) -> sin(x)^3*cos(y);

15. Dec 14, 2007

### jra_1574

OK thanks so much for your help!