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## Homework Statement

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

Integrate f, from x=y....pi/2)

## Homework Equations

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

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- Thread starter jra_1574
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- #1

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

Integrate f, from x=y....pi/2)

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

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- #2

HallsofIvy

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[tex]\int_y^{\pi/2} sin^3(x)cos^3(y) dx[/tex]

Find the anti-derivative (there's a standard method for odd powers of sin or cos) and evaluate between [itex]\pi/2[/itex] and y. Of course, the answer will not be number but will depend on y.

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cristo

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- #4

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

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

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

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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

HallsofIvy

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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.

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

Office_Shredder

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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".

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

cristo

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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

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.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.

- #9

HallsofIvy

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Since cos(y) is a constant (with respect to x)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

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

- #10

Office_Shredder

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I assumed that was what Kreizhn meant. Never mind then

- #11

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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

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

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- #12

HallsofIvy

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

- #13

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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

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[tex] sin^2(x) = \displaystyle\frac{1}{2} \left( 1-cos(2x) \right) [/tex]

then let [itex] u = cos(x) [/itex] so that [itex] du = - sin(x) dx[/itex] 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

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OK thanks so much for your help!

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