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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 innerfunction might be a good idea because the derivative of the innerfunction the cosx is already here but what the outer function? Integration by parts seems more appropriate here...
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 innerfunction might be a good idea because the derivative of the innerfunction the cosx is already here but what the outer function? Integration by parts seems more appropriate here...

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but 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

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.3Woops, my bad. \[\large \int\limits sinxcosx \] Let u = \(cosx\) ; \(\frac{du}{sinx} \)

razor99
 4 years ago
Best ResponseYou've already chosen the best response.0could u tell the question in numbers

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.3You can use both or either sinx or cosx for the substitution btw.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0integration 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
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