anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
razor99
  • razor99
srry i didt get it
anonymous
  • 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
Mimi_x3
  • Mimi_x3
Woops, my bad. \[\large \int\limits sinxcosx \] Let u = \(cosx\) ; \(\frac{du}{-sinx} \)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

razor99
  • razor99
could u tell the question in numbers
Mimi_x3
  • Mimi_x3
You can use both or either sinx or cosx for the substitution btw.
dumbcow
  • dumbcow
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
anonymous
  • anonymous
thx now I got it

Looking for something else?

Not the answer you are looking for? Search for more explanations.