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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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...

Mathematics
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srry i didt get it
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
Woops, my bad. \[\large \int\limits sinxcosx \] Let u = \(cosx\) ; \(\frac{du}{-sinx} \)

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Other answers:

could u tell the question in numbers
You can use both or either sinx or cosx for the substitution btw.
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
thx now I got it

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