Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

frx

I'm a bit confused. I thought I'd solved the problem but when running the integral through Wolframalpha the answer is totally different, am I wrong or is it just the wolframalpha algorithm that doesn't process things like me? http://www.wolframalpha.com/input/?i=integrate+sin^3(x)cos(x) \[\int\limits_{}^{}\sin ^{3}(x) \cos(x) dx\] \[\int\limits_{}^{}(1-\cos ^{2}(x))\cos(x)\sin(x) dx\] \[u= \cos(x); -du=\sin(x) dx\] \[-\int\limits_{}^{}(1-u ^{2})u du= -\int\limits_{}^{}u-u ^{3} du=-\frac{ u ^{2} }{ 2 }+\frac{ u ^{4} }{ 4} +C\] \[=-\frac{\cos ^{2}(x) }{ 2 }+\frac{\cos ^{4}(x) }{ 4}+C\]

  • one year ago
  • one year ago

  • This Question is Closed
  1. hba
    Best Response
    You've already chosen the best response.
    Medals 0

    @frx ake the integral: integral sin^3(x) cos(x) dx For the integrand sin^3(x) cos(x), substitute u = sin(x) and du = cos(x) dx: = integral u^3 du The integral of u^3 is u^4/4: = u^4/4+constant Substitute back for u = sin(x): Answer: | | = (sin^4(x))/4+constant

    • one year ago
  2. hba
    Best Response
    You've already chosen the best response.
    Medals 0

    Take*

    • one year ago
  3. hba
    Best Response
    You've already chosen the best response.
    Medals 0

    That is what wolf says :)

    • one year ago
  4. frx
    Best Response
    You've already chosen the best response.
    Medals 0

    Yepp, that's it but why doesn't the method I used work? I simply just used the fact that sin^2(x)=1-cos^2(x), shouldn't that work, too?

    • one year ago
  5. frx
    Best Response
    You've already chosen the best response.
    Medals 0

    http://www.wolframalpha.com/input/?i=%28sin^4%28x%29%29%2F4%3D-cos^2%28x%29%2F2%2Bcos^4%28x%29%2F4 They do look quite the same?

    • one year ago
  6. frx
    Best Response
    You've already chosen the best response.
    Medals 0

    http://www.wolframalpha.com/input/?i=+sin^4+x++%2F4%3D+-cos^2+x+%2F2%2Bcos^4+x+%2F4

    • one year ago
  7. hartnn
    Best Response
    You've already chosen the best response.
    Medals 1

    both the expressions are equivalent, \[sin^4=[1-cos^2]^2=[1-2cos^2+cos^4]\]

    • one year ago
  8. hartnn
    Best Response
    You've already chosen the best response.
    Medals 1

    the 1/4 which we get extra here, is combined with +c .

    • one year ago
  9. frx
    Best Response
    You've already chosen the best response.
    Medals 0

    So both answers are equally right?

    • one year ago
  10. hartnn
    Best Response
    You've already chosen the best response.
    Medals 1

    exactly.

    • one year ago
  11. frx
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks a lot, both of you! :)

    • one year ago
  12. hartnn
    Best Response
    You've already chosen the best response.
    Medals 1

    welcome ^_^

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.