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Math Help - indefinite integrals

  1. #1
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    indefinite integrals

    hi

    can some one check this please

    i have h(u)=sin^2 (3/4u)

    i used sin^2 u = 1/2(1-cos(2u))

    i made p=3/4u and sin^2 (p) = 1/2(1-cos(2p))

    =1/2(1-cos(3/2u))

    =1/2-1/2cos(3/2u))

    integate=1/2u-3/4sin(3/2u)+c

    many thanks

    wayne
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by smartcar29 View Post
    hi

    can some one check this please

    i have h(u)=sin^2 (3/4u)

    i used sin^2 u = 1/2(1-cos(2u))

    i made p=3/4u and sin^2 (p) = 1/2(1-cos(2p))

    =1/2(1-cos(3/2u))

    =1/2-1/2cos(3/2u))....correct till here

    integate=1/2u-3/4sin(3/2u)+c

    many thanks

    wayne
    your answer is incorrect...

    \int (\frac{1}{2} - \frac{1}{2}\cos{\frac{3u}{2}})du=\frac{1u}{2} - \frac{1}{3}\sin{\frac{3u}{2}}+C
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  3. #3
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    many thanks to you, of course its 1/u so needed to flip it. thanks again
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  4. #4
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    Quote Originally Posted by smartcar29 View Post
    many thanks to you, of course its 1/u so needed to flip it. thanks again
    Are you sure you got the point, here? 'Flipping' is involved, maybe, but in order to cancel the three-over-two fraction that comes as the derivative of 'three over two u', which is the inner function of a chain rule process.

    Maybe you did entirely get the point - then forgive me this diagram opportunity...



    As usual, straight continuous lines differentiate downwards (/integrate up), the straight dashed line similarly but with respect to the dashed balloon expression - so that the triangular network on the right satisfies the chain rule for differentiation.

    The 'flipping' would be, perhaps, the multiplying of the one-over-two fraction by the reciprocal of three-over-two.

    Don't integrate - balloontegrate!

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  5. #5
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    thanks for the cool diagram, to be honest i do not fully understand but i have printed it off and will give it some careful study. many thanks for taking the time.

    wayne
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