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    need to use the substitution x=sin theta
    to integrate 1/(1-x^2)^(3/2)

    ive worked out dx = dtheta cos theta
    and subbed values in but i cant seem to get the right answer which is x/(1-x^2)^(1/2)


    ive got 1/cos^2 theta dtheta so far...
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    and 1/cos^2 theta = sec^2 theta , when u integrate that u get tan theta, no?
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    [QUOTE=cj_134;30583041]and 1/cos^2 theta = sec^2 theta , when u integrate that u get tan theta, no?[/QU
    yh .. but thats not the answer ..
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    yes, then replace ur theta with your x, not directly, notice tan theta = sin theta / cos theta , sin theta = x and cos theta = (1-x^2)^1/2 , which gives the answer you want in terms of x, and dont forget your constant
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    (Original post by cj_134)
    yes, then replace ur theta with your x, not directly, notice tan theta = sin theta / cos theta , sin theta = x and cos theta = (1-x^2)^1/2 , which gives the answer you want in terms of x, and dont forget your constant
    why is cos theta = (1-x^2)^1/2?
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    (Original post by Thisisj)
    why is cos theta = (1-x^2)^1/2?
    sin^2(\theta) + cos^2(\theta) = 1
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    (Original post by Thisisj)
    why is cos theta = (1-x^2)^1/2?
    ok, you know sin^2 theta + cos^2 theta = 1
    so cos^2 theta = 1-sin^2 theta
    cos theta = (1 - sin^2 theta) ^1/2
    you know sin theta = x
    hence cos theta = (1-x^2)^1/2
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    (Original post by EEngWillow)
    sin^2(\theta) + cos^2(\theta) = 1

    (Original post by cj_134)
    ok, you know sin^2 theta + cos^2 theta = 1
    so cos^2 theta = 1-sin^2 theta
    cos theta = (1 - sin^2 theta) ^1/2
    you know sin theta = x
    hence cos theta = (1-x^2)^1/2
    oooooo right...
    thanks guys
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    (Original post by cj_134)
    ok, you know sin^2 theta + cos^2 theta = 1
    so cos^2 theta = 1-sin^2 theta
    cos theta = (1 - sin^2 theta) ^1/2
    you know sin theta = x
    hence cos theta = (1-x^2)^1/2
    do you know any good websites to help me or show a step by step method because im stuck on a few of these types of questions???
 
 
 
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