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UnkleRhaukus

  • 3 years ago

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

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

  2. klimenkov
    • 3 years ago
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    http://en.wikipedia.org/wiki/Beta_function

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

  4. klimenkov
    • 3 years ago
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    Is it a text from a book?

  5. UnkleRhaukus
    • 3 years ago
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    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

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

  7. UnkleRhaukus
    • 3 years ago
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    opsi

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

  9. UnkleRhaukus
    • 3 years ago
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    im not sure how to simplify all that trig

  10. hartnn
    • 3 years ago
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    \(\2sin^2x=1-\cos2x\)

  11. hartnn
    • 3 years ago
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    \(2\cos^2x=1+\cos2x\)

  12. hartnn
    • 3 years ago
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    \(\sin2x=2\sin x\cos x\)

  13. hartnn
    • 3 years ago
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    so u get \(\large \int_0^\pi2\frac{\sin^2\theta}{\sqrt{\cos2\theta}}d\theta\)

  14. hartnn
    • 3 years ago
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    not sure how to proceed further with that.....

  15. UnkleRhaukus
    • 3 years ago
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    looks good

  16. hartnn
    • 3 years ago
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    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

  17. hartnn
    • 3 years ago
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    so that we can use beta function

  18. hartnn
    • 3 years ago
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    i think splitting will help, in 2nd integral, we can put theta = alpha - pi/2

  19. hartnn
    • 3 years ago
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    \(\large \int_0^{\pi/2}2\frac{\sin^2\theta}{\sqrt{\cos2\theta}}d\theta+\int_0^{\pi/2}2\frac{\cos^2\theta}{\sqrt{-cos2\theta}}d\theta\) oh dear....

  20. hartnn
    • 3 years ago
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    i think that didn't help much...:(

  21. UnkleRhaukus
    • 3 years ago
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    its almost in the form of the Beta function !

  22. hartnn
    • 3 years ago
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    1)what about 2theta ? 2) what about minus sign inside sqrt in 2nd integral....imaginary!

  23. hartnn
    • 3 years ago
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    i may have done small/sign mistake, viewers plz verify...

  24. UnkleRhaukus
    • 3 years ago
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    \[2\int\limits^\pi_0\frac{\sin^2(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\]

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

  26. hartnn
    • 3 years ago
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    yeah, as i said, now we have to deal with cos 4x

  27. Eda2012
    • 3 years ago
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    \[x = \sin \theta \] \[\int\limits_{}^{}\frac{ \sqrt{1-\sin ^{2}}\theta }{ \sqrt{1+\sin ^{2}\theta} } . \cos \theta d \theta \] \[\int\limits_{}^{}\frac{ \cos ^{2}\theta }{ \sqrt{\frac{ 2-(1-\cos2\theta) }{ 2 }} }\] \[\int\limits_{}^{}\frac{ \cos ^{2}\theta }{ \sin \theta } d \theta \] \[\int\limits_{}^{} \cot \theta \cos \theta d \theta \]

  28. hartnn
    • 3 years ago
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    how is \(\sqrt{1+\sin^2 \theta}=\sin \theta\) @Eda2012

  29. Eda2012
    • 3 years ago
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    @hartnn sorry...careless mistakes

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

  31. mahmit2012
    • 3 years ago
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    |dw:1352086256918:dw|

  32. mahmit2012
    • 3 years ago
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    |dw:1352086377803:dw|

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