Question

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...

Answer 1

srry i didt get it

Answer 2

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

Answer 3

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

Answer 4

could u tell the question in numbers

Answer 5

You can use both or either sinx or cosx for the substitution btw.

Answer 6

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

Answer 7

thx now I got it