## 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...

1. razor99

srry i didt get it

2. anonymous

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

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

4. razor99

could u tell the question in numbers

5. Mimi_x3

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

6. anonymous

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

thx now I got it