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

UnkleRhaukus

\[\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\]

  • one year ago
  • one year ago

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

    \[\begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&=\int\limits_0^1{(1-x^2)}^{1/2}{(1+x^2)}^{-1/2}\cdot\text dx\\ \\&\text{let}\qquad x^2=u;\qquad x=u^{1/2},\qquad\text d x=\frac 12u^{-1/2}\cdot\text du\\ \\&=\int\limits_0^1{(1-u)}^{1/2}{\left(1+u \right)^{-1/2}}\cdot\frac 12 u ^{-1/2}\cdot\text du\\ \\&=\frac12\int\limits_0^1{(1-u)}^{1/2}{\left(u(1+u) \right)^{-1/2}}\cdot\text du\\ \\&=\frac12\int\limits_0^1{\frac{(1-u)^{1/2}}{\left(u+u^2 \right)^{1/2}}}\cdot\text du\\ \\&\\ \\&\\ \\&=\qquad\color{red}{???}\\ \\&\\ \\&=\frac{1}{2}\left(\frac{\text B\left(\frac12,\frac14\right)}{2}-\text B\left(-\frac12,\frac34\right)\right)\\ \\&=\frac{1}{4}\text B\left(\frac12,\frac14\right)-\frac{1}{2}\text B\left(-\frac12,\frac34\right)\\ \\&=\frac{1}{4}\left(\frac{\Gamma\left(\frac12\right)\Gamma \left(\frac14\right)}{\Gamma\left(\frac34\right)}\right)-\left(\frac{\Gamma\left(\frac12\right)\Gamma \left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)\\ \\&=\frac{1}{4}\sqrt\pi\left(\frac{\Gamma \left(\frac14\right)}{\Gamma\left(\frac34\right)}-4\frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)\\ \\&\approx0.712\\ \end{align*}\]

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

    http://en.wikipedia.org/wiki/Beta_function

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

    \[\text B(m,n)=\int\limits_0^1x^{m-1}(1-x)^{n-1} \cdot\text d x\]\[\qquad=\int\limits_0^\infty \frac{y^{n-1}}{(1+y)^{m+n}}\cdot\text d y\]\[\quad=2\int\limits_0^{\pi/2} \cos^{2m-1}(\theta) \sin^{2n-1}(\theta)\cdot\text d \theta\]

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

    Is it a text from a book?

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

    the first line is from a book \[\begin{align*}\\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&=\frac{1}{4}\sqrt\pi\left(\frac{\Gamma \left(\frac14\right)}{\Gamma\left(\frac34\right)}-4\frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)\\\end{align*} \]and this is answer at the back of the book

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

    \[\begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&x^2=\cos(2\theta)\qquad\text dx=\frac{-\sin(2\theta)}x\text d\theta\\ \\&x=0\rightarrow\theta=\pi\qquad x=1\rightarrow\theta=0\\ \\&\int\limits_\pi^0\sqrt{\frac{1-\cos^2(2\theta)}{1+\cos^2(2\theta)}}\cdot\frac{-\sin(2\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \end{align*}\]

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

    opsi

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

    \[\begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&x^2=\cos(2\theta)\\ \\&\text dx=\frac{-\sin(2\theta)}x\text d\theta\\ \\&x=0\rightarrow\theta=\pi\qquad x=1\rightarrow\theta=0\\ \\&\int\limits_\pi^0\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}\cdot\frac{-\sin(2\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \end{align*}\]

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

    im not sure how to simplify all that trig

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

    \(\2sin^2x=1-\cos2x\)

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

    \(2\cos^2x=1+\cos2x\)

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

    \(\sin2x=2\sin x\cos x\)

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

    so u get \(\large \int_0^\pi2\frac{\sin^2\theta}{\sqrt{\cos2\theta}}d\theta\)

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

    not sure how to proceed further with that.....

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

    looks good

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

    to get limits 0 to pi/2, we have put theta = 2x and have to deal with sqrt(cos 4x) in denominator..... or we can try splitting integral from 0 tp pi/2 and pi/2 to pi and some how convert 2nd integral also to 0 to pi/2

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

    so that we can use beta function

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

    i think splitting will help, in 2nd integral, we can put theta = alpha - pi/2

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

    \(\large \int_0^{\pi/2}2\frac{\sin^2\theta}{\sqrt{\cos2\theta}}d\theta+\int_0^{\pi/2}2\frac{\cos^2\theta}{\sqrt{-cos2\theta}}d\theta\) oh dear....

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

    i think that didn't help much...:(

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

    its almost in the form of the Beta function !

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

    1)what about 2theta ? 2) what about minus sign inside sqrt in 2nd integral....imaginary!

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

    i may have done small/sign mistake, viewers plz verify...

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

    \[2\int\limits^\pi_0\frac{\sin^2(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\]

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

    \[\begin{align*} \\&\int\limits_\pi^0\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}\cdot\frac{-\sin(2\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \\&2\sin^2(x)=1-\cos(2x)&2\cos^2(x)=1+\cos(2x)\\\\&\sin(2x)=2\sin(x)\cos(x)&\cos(2x)=\cos^2 (x)-\sin^2(x)\\ \\&\int\limits_\pi^0\sqrt{\frac{2\sin^2\theta }{2\cos^2\theta}}\cdot\frac{-2\sin(\theta)\cos(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \\&2\int\limits^\pi_0\tan(\theta)\cdot\frac{\sin(\theta)\cos(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \\&2\int\limits^\pi_0\frac{\sin^2(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \\&\text{let } \theta=2\phi\qquad\text d\theta =2\text d\phi\\ \\&\theta=0\rightarrow\phi=0\qquad\theta=\pi\rightarrow\phi=\pi/2\\ \\&2\int\limits_0^{\pi/2}\frac{\sin^2(2\phi)}{\sqrt{\cos(4\phi)}}2\text d\phi\\ \\&4\int\limits_0^{\pi/2}\frac{\sin^2(2\phi)}{\sqrt{\cos(4\phi)}}\text d\phi\\\end{align*}\]

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

    yeah, as i said, now we have to deal with cos 4x

    • one year ago
  27. Eda2012
    Best Response
    You've already chosen the best response.
    Medals 0

    \[x = \sin \theta \] \[\int\limits_{}^{}\frac{ \sqrt{1-\sin ^{2}}\theta }{ \sqrt{1+\sin ^{2}\theta} } . \cos \theta d \theta \] \[\int\limits_{}^{}\frac{ \cos ^{2}\theta }{ \sqrt{\frac{ 2-(1-\cos2\theta) }{ 2 }} }\] \[\int\limits_{}^{}\frac{ \cos ^{2}\theta }{ \sin \theta } d \theta \] \[\int\limits_{}^{} \cot \theta \cos \theta d \theta \]

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

    how is \(\sqrt{1+\sin^2 \theta}=\sin \theta\) @Eda2012

    • one year ago
  29. Eda2012
    Best Response
    You've already chosen the best response.
    Medals 0

    @hartnn sorry...careless mistakes

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

    \[\begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&=\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\times\sqrt{\frac{1-x^2}{1-x^2}}\cdot\text dx\\ \\&=\int\limits_0^1\frac{1-x^2}{\sqrt{1-x^4}}\cdot\text dx\\ \\&=\int\limits_0^1(1-x^4)^{-1/2}\cdot\text dx-\int\limits_0^1x^2(1-x^4)^{-1/2}\text dx\\ \\\text{let }x=u^{1/4}\qquad&\text dx=\frac14 u^{-3/4}\text du\\ \\&\small =\int\limits_0^1(1-u)^{-1/2}\cdot\frac14 u^{-3/4}\text du-\int\limits_0^1u^{1/2}(1-u)^{-1/2}\cdot\frac14 u^{-3/4}\text du\\ \\&=\frac14\int\limits_0^1u^{-3/4}(1-u)^{-1/2}\cdot \text du-\frac14\int\limits_0^1u^{-1/4}(1-u)^{-1/2}\cdot\text du\\ \\\text B(m,n)=\int\limits_0^1t^{m-1}(1-t)^{n-1} \cdot\text d t\\ \\m_1-1=-3/4\qquad& n_1-1=-1/2&\qquad \qquad\qquad m_2-1=-1/4\qquad n_2-1=-1/2\\ \\m_1=1/4\qquad \qquad & n_1=1/2&\qquad\qquad\qquad m_2=3/4\qquad\qquad n_2=1/2\\ \\&=\frac14\text B\left(\frac 14,\frac12\right)-\frac14\text B\left(\frac34,\frac12\right)\\ \\&=\frac14\text B\left(\frac 14,\frac12\right)-\frac14\text B\left(\frac34,\frac12\right)\\ \\&=\frac 14\left(\frac{\Gamma(\frac 14)\Gamma(\frac 12)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)\Gamma(\frac 12)}{\Gamma(\frac 54)}\right)\\ \\&=\frac {\sqrt \pi}4\left(\frac{\Gamma(\frac 14)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)}{\Gamma(\frac 54)}\right)\\ \\&=\frac {\sqrt \pi}4\left(\frac{\Gamma(\frac 14)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)}{\frac 14\Gamma(\frac 14)}\right)\\ \\&=\frac {\sqrt \pi}4\left(\frac{\Gamma(\frac 14)}{\Gamma(\frac 34)}-4\frac{\Gamma(\frac34)}{\Gamma(\frac 14)}\right)\\ \end{align*}\]

    • one year ago
  31. mahmit2012
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1352086256918:dw|

    • one year ago
  32. mahmit2012
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1352086377803:dw|

    • 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.