anonymous
  • anonymous
Help with following integral
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{0}^{\pi/2} e ^{sint}cost dt\]
anonymous
  • anonymous
sin t* I can't seem to figure out what to use. everytime I get e I get lost x.x
RadEn
  • RadEn
use integral by u-sub let u = sint du = cost dt

Looking for something else?

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

More answers

anonymous
  • anonymous
but then I get stuck with \[e ^{u}\] wouldn't that require integration by part?
RadEn
  • RadEn
just still it self int e^u = e^u
anonymous
  • anonymous
Oh only e^2u would require by part? ( question by curiosity)
RadEn
  • RadEn
e^u not e^2u now, subtitute back that u = sint so, it is [e^(sint)]
anonymous
  • anonymous
Thank you :)
RadEn
  • RadEn
the rest is calculation difference of limits, can u ? you're welcome :)
anonymous
  • anonymous
Yeah I can I get confuse with e. I know sometimes it requires integration by part but this one was a simple case. Thanks a lot though helped me understand a bit more!
RadEn
  • RadEn
|dw:1366095285032:dw|
RadEn
  • RadEn
that's the last answer

Looking for something else?

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