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    Hi guys, stuck on integration here, take a look pls

    Integrate from a to 0 the function (a^2 - a^2(sinx)^2)^1/2

    There is a hint given, it says to use the subsitution y = asin(theta)

    If anyone knows any other methods to integrate this function that arent too complex id love to know of them too, thanks
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    (Original post by karnten07)
    Hi guys, stuck on integration here, take a look pls

    Integrate from a to 0 the function (a^2 - a^2(sinx)^2)^1/2

    There is a hint given, it says to use the subsitution y = asin(theta)

    If anyone knows any other methods to integrate this function that arent too complex id love to know of them too, thanks
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle \int_a^0{ \sqrt{a^2 - a^2 \sin^2 x} \ \mathrm{d}x


    \displaystyle = \int_a^0{ a \cos x} \ \mathrm{d}x

    You could perform a substitution now but it's simpler to just integrate directly.

    \displaystyle = \left[ a \sin x \right]_a^0

    \displaystyle = -a \sin a

    I have my doubts about this question. The limits are very odd, and the substitution hint involves y and \theta but not x which is the variable in the question.
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    \int_a^0 \sqrt{a^2 - a^2 sin^2x},dx



= a \int_a^0 \sqrt{1 - sin^2x},dx

    But 1 - sin²x = cos²x, and I'll let you continue from here ...

    I don't understand why they've said to use that substitution. Are you sure that's the correct question?
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    Lol :p: , no replies for six hours then two within three minutes!
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    (Original post by karnten07)
    Hi guys, stuck on integration here, take a look pls

    Integrate from a to 0 the function (a^2 - a^2(sinx)^2)^1/2

    There is a hint given, it says to use the subsitution y = asin(theta)

    If anyone knows any other methods to integrate this function that arent too complex id love to know of them too, thanks
    Sorry guys, it was late and i wrote it out wrong. I wrote out the question out having already carried out the substitution. It shud read

    Integrate from a to 0 the function (a^2 - y^2)^1/2 using y = asin(theta) as a substitution.

    So i get to a^3 INT (from pi/2 to 0) d(theta) (cos(Thetha))^2.(1-sintheta)

    I know that part is right cos it says it in the solution, but im getting stuck on this integration by parts from here, any help guys?
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    (Original post by karnten07)
    Sorry guys, it was late and i wrote it out wrong. I wrote out the question out having already carried out the substitution. It shud read

    Integrate from a to 0 the function (a^2 - y^2)^1/2 using y = asin(theta) as a substitution.

    So i get to a^3 INT (from pi/2 to 0) d(theta) (cos(Thetha))^2.(1-sintheta)

    I know that part is right cos it says it in the solution, but im getting stuck on this integration by parts from here, any help guys?
    It's not correct, or the question is still wrong:

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle \int_a^0{\sqrt{a^2 - y^2}\ \mathrm{d}y = \int_{\frac{\pi}{2}}^0{a^2 \cos^2 \theta}\ \mathrm{d}\theta


    There's no integration by parts here, so I'm not sure what you are stuck on.
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    (Original post by karnten07)
    I know that part is right cos it says it in the solution, but im getting stuck on this integration by parts from here, any help guys?
    What are you doing?
    dy/dt=acos(t) so dy=acos(t)dt therefore INT acos(t)*acos(t)dt=INTa^2cos^2(t) dt

    edit: Sorry should press refresh before posting
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    (Original post by karnten07)
    Sorry guys, it was late and i wrote it out wrong. I wrote out the question out having already carried out the substitution. It shud read

    Integrate from a to 0 the function (a^2 - y^2)^1/2 using y = asin(theta) as a substitution.

    So i get to a^3 INT (from pi/2 to 0) d(theta) (cos(Thetha))^2.(1-sintheta)

    I know that part is right cos it says it in the solution, but im getting stuck on this integration by parts from here, any help guys?
    Oh my, im really sorry, the integration is the same as before except it should be multiplied by (a-y). so im hoping it should come to the point where im up to? :confused: :rolleyes:
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    I think the reason i was getting confused is because this is the second part of a double integral, sorry again.
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    (Original post by karnten07)
    I think the reason i was getting confused is because this is the second part of a double integral, sorry again.
    Perhaps you should post the original question and where you think you've got to. This way we can be sure that we're all looking at the same problem.
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    (Original post by Mathmoid)
    Perhaps you should post the original question and where you think you've got to. This way we can be sure that we're all looking at the same problem.
    Ok, so im asked to evaluate the double integral:

    ~~D sqrt(a^2 - y^2) dA where D is the triangle with vertices (0,0) (a,0) (a,a)

    Ive Put the two squiggles to represent the integration symbol twice.

    So then i integrated with respect to x from a to 0 which is where the (a-y) comes in that i had forgot to write in my original question. Then i integrate with respect to y from a to y and using the substitution y = asintheta.
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    (Original post by karnten07)
    Oh my, im really sorry, the integration is the same as before except it should be multiplied by (a-y). so im hoping it should come to the point where im up to? :confused: :rolleyes:
    Okay, well I get the same as you, but don't do parts on that whole thing...

    hint:
    Spoiler:
    Show
    \int a^3\cos^2(\theta)d\theta-\int a^3\sin(\theta)d\theta
    The second one is simple, and for the first one, use cos(2x)=2cos^2(x)-1 to change the integral into something easier
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    (Original post by nota bene)
    Okay, well I get the same as you, but don't do parts on that whole thing...

    hint:
    Spoiler:
    Show
    \int a^3\cos^2(\theta)d\theta-\int a^3\sin(\theta)d\theta
    The second one is simple, and for the first one, use cos(2x)=2cos^2(x)-1 to change the integral into something easier
    mmm my integration sucks. ill come back to this later, thanks for the help guys, im tired aswell, so gonna have a rest and then pick it up after. :cool:
 
 
 

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