Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Calculus1
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

\[\begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&=\int\limits_0^1{(1-x^2)}^{1/2}{(1+x^2)}^{-1/2}\cdot\text dx\\ \\&\text{let}\qquad x^2=u;\qquad x=u^{1/2},\qquad\text d x=\frac 12u^{-1/2}\cdot\text du\\ \\&=\int\limits_0^1{(1-u)}^{1/2}{\left(1+u \right)^{-1/2}}\cdot\frac 12 u ^{-1/2}\cdot\text du\\ \\&=\frac12\int\limits_0^1{(1-u)}^{1/2}{\left(u(1+u) \right)^{-1/2}}\cdot\text du\\ \\&=\frac12\int\limits_0^1{\frac{(1-u)^{1/2}}{\left(u+u^2 \right)^{1/2}}}\cdot\text du\\ \\&\\ \\&\\ \\&=\qquad\color{red}{???}\\ \\&\\ \\&=\frac{1}{2}\left(\frac{\text B\left(\frac12,\frac14\right)}{2}-\text B\left(-\frac12,\frac34\right)\right)\\ \\&=\frac{1}{4}\text B\left(\frac12,\frac14\right)-\frac{1}{2}\text B\left(-\frac12,\frac34\right)\\ \\&=\frac{1}{4}\left(\frac{\Gamma\left(\frac12\right)\Gamma \left(\frac14\right)}{\Gamma\left(\frac34\right)}\right)-\left(\frac{\Gamma\left(\frac12\right)\Gamma \left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)\\ \\&=\frac{1}{4}\sqrt\pi\left(\frac{\Gamma \left(\frac14\right)}{\Gamma\left(\frac34\right)}-4\frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)\\ \\&\approx0.712\\ \end{align*}\]
http://en.wikipedia.org/wiki/Beta_function
\[\text B(m,n)=\int\limits_0^1x^{m-1}(1-x)^{n-1} \cdot\text d x\]\[\qquad=\int\limits_0^\infty \frac{y^{n-1}}{(1+y)^{m+n}}\cdot\text d y\]\[\quad=2\int\limits_0^{\pi/2} \cos^{2m-1}(\theta) \sin^{2n-1}(\theta)\cdot\text d \theta\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Is it a text from a book?
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
\[\begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&x^2=\cos(2\theta)\qquad\text dx=\frac{-\sin(2\theta)}x\text d\theta\\ \\&x=0\rightarrow\theta=\pi\qquad x=1\rightarrow\theta=0\\ \\&\int\limits_\pi^0\sqrt{\frac{1-\cos^2(2\theta)}{1+\cos^2(2\theta)}}\cdot\frac{-\sin(2\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \end{align*}\]
opsi
\[\begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&x^2=\cos(2\theta)\\ \\&\text dx=\frac{-\sin(2\theta)}x\text d\theta\\ \\&x=0\rightarrow\theta=\pi\qquad x=1\rightarrow\theta=0\\ \\&\int\limits_\pi^0\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}\cdot\frac{-\sin(2\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \end{align*}\]
im not sure how to simplify all that trig
\(\2sin^2x=1-\cos2x\)
\(2\cos^2x=1+\cos2x\)
\(\sin2x=2\sin x\cos x\)
so u get \(\large \int_0^\pi2\frac{\sin^2\theta}{\sqrt{\cos2\theta}}d\theta\)
not sure how to proceed further with that.....
looks good
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
so that we can use beta function
i think splitting will help, in 2nd integral, we can put theta = alpha - pi/2
\(\large \int_0^{\pi/2}2\frac{\sin^2\theta}{\sqrt{\cos2\theta}}d\theta+\int_0^{\pi/2}2\frac{\cos^2\theta}{\sqrt{-cos2\theta}}d\theta\) oh dear....
i think that didn't help much...:(
its almost in the form of the Beta function !
1)what about 2theta ? 2) what about minus sign inside sqrt in 2nd integral....imaginary!
i may have done small/sign mistake, viewers plz verify...
\[2\int\limits^\pi_0\frac{\sin^2(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\]
\[\begin{align*} \\&\int\limits_\pi^0\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}\cdot\frac{-\sin(2\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \\&2\sin^2(x)=1-\cos(2x)&2\cos^2(x)=1+\cos(2x)\\\\&\sin(2x)=2\sin(x)\cos(x)&\cos(2x)=\cos^2 (x)-\sin^2(x)\\ \\&\int\limits_\pi^0\sqrt{\frac{2\sin^2\theta }{2\cos^2\theta}}\cdot\frac{-2\sin(\theta)\cos(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \\&2\int\limits^\pi_0\tan(\theta)\cdot\frac{\sin(\theta)\cos(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \\&2\int\limits^\pi_0\frac{\sin^2(\theta)}{\sqrt{\cos(2\theta)}}\text d\theta\\ \\&\text{let } \theta=2\phi\qquad\text d\theta =2\text d\phi\\ \\&\theta=0\rightarrow\phi=0\qquad\theta=\pi\rightarrow\phi=\pi/2\\ \\&2\int\limits_0^{\pi/2}\frac{\sin^2(2\phi)}{\sqrt{\cos(4\phi)}}2\text d\phi\\ \\&4\int\limits_0^{\pi/2}\frac{\sin^2(2\phi)}{\sqrt{\cos(4\phi)}}\text d\phi\\\end{align*}\]
yeah, as i said, now we have to deal with cos 4x
\[x = \sin \theta \] \[\int\limits_{}^{}\frac{ \sqrt{1-\sin ^{2}}\theta }{ \sqrt{1+\sin ^{2}\theta} } . \cos \theta d \theta \] \[\int\limits_{}^{}\frac{ \cos ^{2}\theta }{ \sqrt{\frac{ 2-(1-\cos2\theta) }{ 2 }} }\] \[\int\limits_{}^{}\frac{ \cos ^{2}\theta }{ \sin \theta } d \theta \] \[\int\limits_{}^{} \cot \theta \cos \theta d \theta \]
how is \(\sqrt{1+\sin^2 \theta}=\sin \theta\) @Eda2012
@hartnn sorry...careless mistakes
\[\begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\cdot\text dx\\ \\&=\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\times\sqrt{\frac{1-x^2}{1-x^2}}\cdot\text dx\\ \\&=\int\limits_0^1\frac{1-x^2}{\sqrt{1-x^4}}\cdot\text dx\\ \\&=\int\limits_0^1(1-x^4)^{-1/2}\cdot\text dx-\int\limits_0^1x^2(1-x^4)^{-1/2}\text dx\\ \\\text{let }x=u^{1/4}\qquad&\text dx=\frac14 u^{-3/4}\text du\\ \\&\small =\int\limits_0^1(1-u)^{-1/2}\cdot\frac14 u^{-3/4}\text du-\int\limits_0^1u^{1/2}(1-u)^{-1/2}\cdot\frac14 u^{-3/4}\text du\\ \\&=\frac14\int\limits_0^1u^{-3/4}(1-u)^{-1/2}\cdot \text du-\frac14\int\limits_0^1u^{-1/4}(1-u)^{-1/2}\cdot\text du\\ \\\text B(m,n)=\int\limits_0^1t^{m-1}(1-t)^{n-1} \cdot\text d t\\ \\m_1-1=-3/4\qquad& n_1-1=-1/2&\qquad \qquad\qquad m_2-1=-1/4\qquad n_2-1=-1/2\\ \\m_1=1/4\qquad \qquad & n_1=1/2&\qquad\qquad\qquad m_2=3/4\qquad\qquad n_2=1/2\\ \\&=\frac14\text B\left(\frac 14,\frac12\right)-\frac14\text B\left(\frac34,\frac12\right)\\ \\&=\frac14\text B\left(\frac 14,\frac12\right)-\frac14\text B\left(\frac34,\frac12\right)\\ \\&=\frac 14\left(\frac{\Gamma(\frac 14)\Gamma(\frac 12)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)\Gamma(\frac 12)}{\Gamma(\frac 54)}\right)\\ \\&=\frac {\sqrt \pi}4\left(\frac{\Gamma(\frac 14)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)}{\Gamma(\frac 54)}\right)\\ \\&=\frac {\sqrt \pi}4\left(\frac{\Gamma(\frac 14)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)}{\frac 14\Gamma(\frac 14)}\right)\\ \\&=\frac {\sqrt \pi}4\left(\frac{\Gamma(\frac 14)}{\Gamma(\frac 34)}-4\frac{\Gamma(\frac34)}{\Gamma(\frac 14)}\right)\\ \end{align*}\]
|dw:1352086256918:dw|
|dw:1352086377803:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question