Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Is it a text from a book?

Answer 5

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

Answer 6

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

Answer 7

opsi

Answer 8

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

Answer 9

im not sure how to simplify all that trig

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

looks good

Answer 16

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

Answer 17

so that we can use beta function

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

its almost in the form of the Beta function !

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

@hartnn sorry...careless mistakes

Answer 30

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