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    Is my working right so far?


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    (Original post by SDavis123)
    Is my working right so far?


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    I think integration by parts is the way to go with this one
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    Yes. That is correct.

    Now you have to put it back in terms of 'x'.
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    (Original post by claret_n_blue)
    Yes. That is correct.

    Now you have to put it back in terms of 'x'.
    But if I do that I get

    Sin^(-1)x.sin.sin^(-1) + cos.sin^(-1) + k

    How would you simplify that?


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    (Original post by SDavis123)
    But if I do that I get

    Sin^(-1)x.sin.sin^(-1) + cos.sin^(-1) + k

    How would you simplify that?
    For the

     \sin( \sin^{-1}(x))

    that uses a similar principle to  \sqrt{x^2} . So I'm sure you can figure out what that is.

    For the second bit, you need to remember this identity:

     \sin(\arccos(x)) = \cos(\arcsin(x)) = \sqrt{1 - x^2} .

    I'm just working through the proof now, I will edit this post and write it up when I have done.


    ................


    PROOF:

    'O' = Opposite
    'H' = Hypotenuse
    'A' = Adjacent

    We know that  \sin = \frac{\mathrm{O}}{\mathrm{H}}

    Now, in a triangle defined by  \sin(t) = x , we have defined  \mathrm{O} = x and  \mathrm{H} = 1 .

    From Pythagoras theorem, we know that  O^2 + A^2 = H^2 . So that means that

     x^2 + A^2 = 1^2 ,

    which can be re-written as

     \mathrm{A}^2 = 1 - x^2 \implies \mathrm{A} = \sqrt{1 - x^2} .

    Now we also know that  \cos = \frac{\mathrm{A}}{\mathrm{H}} . We have already said that  \mathrm{H} = 1, \mathrm{A} = \sqrt{1 - x^2} .

    So in this triangle, we now have  \cos(t) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2} .

    From before, we said that  \sin(t) = x \implies t = \arcsin(x) .

    So we get

     \cos(\arcsin(x)) = \sqrt{1 - x^2} .
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    (Original post by SDavis123)
    But if I do that I get

    Sin^(-1)x.sin.sin^(-1) + cos.sin^(-1) + k

    How would you simplify that?


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    It's quite simple.

    Given

    x=\sin u

    surely you can simplify u\sin u

    by noting that

    u=\arcsin(x)

    Also

    \sin^2 u + \cos^2 u = 1
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    Ok I think I get it but I definitely wouldn't have thought of this myself

    And thank you so much for the help guys, I really appreciate the effort you put in, especially with that proof


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