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anonymous

  • 4 years ago

As a homework I am supposed to integrate sinxcosx using the integration method of substitution. I thought taking sinx as the inner-function might be a good idea because the derivative of the inner-function the cosx is already here but what the outer function? Integration by parts seems more appropriate here...

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  1. razor99
    • 4 years ago
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    srry i didt get it

  2. anonymous
    • 4 years ago
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    but then I do the integral of tdt after replacing sinx with t and cosx with dt and get that the integral of sinxcosx=0.5*x^2

  3. Mimi_x3
    • 4 years ago
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    Woops, my bad. \[\large \int\limits sinxcosx \] Let u = \(cosx\) ; \(\frac{du}{-sinx} \)

  4. razor99
    • 4 years ago
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    could u tell the question in numbers

  5. Mimi_x3
    • 4 years ago
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    You can use both or either sinx or cosx for the substitution btw.

  6. dumbcow
    • 4 years ago
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    integration by parts could work....substitution would be more popular and simpler because the 1 function is derivatives of other u = sin x du = cos x dx --> dx = du/cos x cos x cancels --> integral u du = u^2/2 = sin^2(x)/2

  7. anonymous
    • 4 years ago
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    thx now I got it

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