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    Would appreciate some help with this question I'm finding tricky.

    By using the substitution given, show that the LHS gives the RHS [6]
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    (Original post by latslikeabat)
    Would appreciate some help with this question I'm finding tricky.

    By using the substitution given, show that the LHS gives the RHS [6]
    Whenever you post, explain what you have attempted to do or better, write that what you have done so far, where you have got to so then someone can better give you hints.
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    oops my bad.

    well i got:



    I then used the substition to form



    and hence



    =



    =



    =



    I wasn't sure what to do after this but considered using parts, however that seemed overly complex.
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    (Original post by latslikeabat)
    oops my bad.

    well i got:



    I then used the substition to form



    and hence



    =



    =



    =



    I wasn't sure what to do after this but considered using parts, however that seemed overly complex.
    Ok great, can work with that. You've got the right idea but the notation is a bit sloppy though, in terms of what you're integrating with respect to. After the substitution, you want to be integrating with respect to theta, not a, a is just a constant here.

     \displaystyle \int^{\frac{a}{2}}_a \sqrt{a^2-x^2} \ dx

     x = a \sin \theta

     \dfrac{dx}{d \theta} = a \cos \theta

    so

     dx = a \cos \theta d \theta

    Changing limits and subbing in to put in terms of theta.

     \displaystyle \int ^{\frac{\pi}{6}}_{\frac{\pi}{2}} \sqrt{a^2cos^2\theta}acos \theta \ d \theta
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    (Original post by Clarity Incognito)
    Ok great, can work with that. You've got the right idea but the notation is a bit sloppy though, in terms of what you're integrating with respect to. After the substitution, you want to be integrating with respect to theta, not a, a is just a constant here.

     \displaystyle \int^{\frac{a}{2}}_a \sqrt{a^2-x^2} \ dx

     x = a \sin \theta

     \dfrac{dx}{d \theta} = a \cos \theta

    so

     dx = a \cos \theta d \theta

    Changing limits and subbing in to put in terms of theta.

     \displaystyle \int ^{\frac{\pi}{6}}_{\frac{\pi}{2}} \sqrt{a^2cos^2\theta}acos \theta \ d \theta

    Ah ok , that makes a lot more sense.

    and so after that would you get:



    and then use the double angle formula so that:



    After that point would you use parts so that:



    and



    ?

    Also what is 'u' differentiated w.r.t?
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    this is how you do it
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    (Original post by latslikeabat)
    Ah ok , that makes a lot more sense.

    and so after that would you get:



    and then use the double angle formula so that:



    After that point would you use parts so that:



    and



    ?

    Also what is 'u' differentiated w.r.t?
    Since a is a constant you don't need to use "by parts". What do you get if you integrate 7 cos x? Just 7 sin x. Treat the a^2 like a 7.
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    (Original post by latslikeabat)
    Ah ok , that makes a lot more sense.
    Also what is 'u' differentiated w.r.t?
    a is just a constant so a^2 is also a constant. You can take it out of the integral. You're integrating with respect to theta.

    (Original post by irfan91)
    this is how you do it
    Just a note, do you post full solutions, the point of the forum is to help users gain a better understanding, not to give them solutions. Please read the forum rules.
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    (Original post by Clarity Incognito)
    a is just a constant so a^2 is also a constant. You can take it out of the integral. You're integrating with respect to theta.



    Just a note, do you post full solutions, the point of the forum is to help users gain a better understanding, not to give them solutions. Please read the forum rules.

    lool sorry officer
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    (Original post by irfan91)
    lool sorry officer
    Haha, didn't mean for it to sound so angry, I just wanted to point it out, it's quite usual for members to let newer members know.

    EDIT: Also, I just looked through your working and you don't account for the limits of substitution properly. The top limit should be pi/6 throughout and the bottom pi/2 unless you swap them and put a minus out front. The integral actually gives area under the x axis and as area is positive, you take the magnitude of the answer.
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    Ok, now I get it.

    Thanks for the help.
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    (Original post by Clarity Incognito)
    Haha, didn't mean for it to sound so angry, I just wanted to point it out, it's quite usual for members to let newer members know.

    EDIT: Also, I just looked through your working and you don't account for the limits of substitution properly. The top limit should be pi/6 throughout and the bottom pi/2 unless you swap them and put a minus out front. The integral actually gives area under the x axis and as area is positive, you take the magnitude of the answer.
    its all good, yeah i tried it first time with the limits the other way round and i didn't get the RHS so i just flipped it to get the RHS, thanks for the advice and forum rules
 
 
 
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