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

Callisto

Integration problem... \[\int \frac{dx}{sin^4x+cos^4x}\]

  • one year ago
  • one year ago

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

    \[=\int \frac{sec^4x \ dx}{tan^4x +1}\]\[=\int \frac{(tan^2x+1)sec^2x \ dx}{tan^4x +1}\] \[= \frac{(tan^2x+1) \ d(tanx)}{tan^4x +1}\] Let t=tanx Integral becomes\[= \frac{(t^2+1) \ dt}{t^4 +1}\] How should I continue?

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

    @shivam_bhalla :(

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

    Partial fractions, but first \(t^4+1=t^4+2t^2+1-2t^2=(t^2+1)^2-(t\sqrt{2})^2\)

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

    Why haven't my teacher taught me partial fraction....... /_\

    • one year ago
  5. 1729antony
    Best Response
    You've already chosen the best response.
    Medals 1

    try this divide throught out by t^2 so it becomes (1+1/t^2)/(t^2 +1/t^2).. then substitute t-(1/t) =m so the equation becomes dm/(m^2 +2) which you can integrate

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

    @1729antony , is absolutely correct. :)

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

    @blockcolder , with partial fractions, the problem becomes hectic to solve

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

    I just remembered a problem wherein a t^4+1 came out, and I immediately thought of partial fractions, since that's what I did that time.

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

    So, here it goes~ \[\int \frac{dx}{sin^4x+cos^4x} = \int\frac{sec^4x}{tan^4x+1}dx = \int \frac{tan^2x+1}{tan^4+1}d(tanx) \]\[= \int \frac{t^2+1}{t^4+1}dt = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt \]\[Let \ m=t-\frac{1}{t}, \ dm = 1+\frac{1}{t^2}dt\]\[ \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt = \int \frac{1}{m^2+2}dm = \int \frac{\sqrt2sec^2y}{(\sqrt2tany)^2+2}dy\]\[= \frac{\sqrt2}{2}\int \frac{sec^2y}{tan^2y+1}dy = \frac{\sqrt2}{2} \int dy = \frac{\sqrt{2}}{2}y+C\]\[ = \frac{\sqrt{2}}{2} tan^{-1}(\frac{tanx-\frac{1}{tanx}}{\sqrt2})+C\] Sigh... Probably, it's going to be wrong :|

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

    @Callisto , you got it right. Well done :)

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

    Finally... (burst into tears...)

    • 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.