To show that: $\int_{-1}^{1} \frac{1}{(1+x)^{1/3}\ (1-x)^{2/3}} dx = \frac{2\pi}{\sqrt3}$

Question

To show that:

$$\int_{-1}^{1} \frac{1}{(1+x)^{1/3}\ (1-x)^{2/3}} dx = \frac{2\pi}{\sqrt3}$$...............(A)

We see that $\frac{2\pi}{\sqrt3}$ can be written as $$\frac{\pi}{\sqrt3 /2} = \frac{\pi}{\sin (\pi/3) } = \Gamma(1/3) \Gamma(2/3) = \frac{\Gamma(1/3) \Gamma(2/3)}{\Gamma(1)} = B(1/3,2/3).$$

And, $$B(1/3,2/3) = \int_0^1 x^{1/3-1} (1-x)^{2/3-1} dx$$..........(B)

Finding difficult to find a relation between L.H.S. of (A) and (B)....

Answer 1

By the substitution $u = (1 + x)/2$, we have

$$\int_{-1}^1 \frac{1}{(1 + x)^{1/3}(1 - x)^{2/3}}\, dx = \int_0^1 \frac{1}{(2u)^{1/3}(2 - 2u)^{2/3}}\, (2\,du) = \int_0^1 u^{-1/3} (1 - u)^{-2/3}\, du.$$

Answer 2

HINT: let
$$\int_{-1}^{1}\frac{1}{(1+x)^{1/3}(1-x)^{2/3}}\ dx=\int_{-1}^{1}\frac{(1+x)^{1/3}}{(1+x)^{2/3}(1-x)^{2/3}}\ dx=\int_{-1}^{1}\frac{(1+x)^{1/3}}{(1-x^2)^{2/3}}\ dx$$ now, substitute $x=\cos\theta\implies dx=-\sin \theta\ d\theta$ $$=\int_{\pi}^{0}\frac{(1+\cos\theta)^{1/3}(-\sin \theta \ d\theta)}{(1-\cos^2\theta)^{2/3}}$$ $$=\int_{0}^{\pi}\frac{(2\sin^2\frac{\theta}{2})^{1/3}(\sin \theta \ d\theta)}{\sin^3\theta}$$ $$=2^{2/3}\int_{0}^{\pi}\frac{\sin^{2/3}\left(\frac{\theta}{2}\right)\ d\theta}{\sin^2\theta}$$ $$=2^{2/3}\int_{0}^{\pi}\frac{\sin^{2/3}\left(\frac{\theta}{2}\right)\ d\theta}{4\sin^{2}\left(\frac{\theta}{2}\right)\cos^{2}\left(\frac{\theta}{2}\right)}$$

Answer 3

This is too long for a comment. Depending on the question is to show that the integral equals what it equals, or just to see how one get to one integral into the next, this answer might be offtopic. If so, tell me and I remove it.

First, write the function as $$ \begin{split} \frac{1}{(1+x)^{1/3}(1-x)^{2/3}}&=\frac{(1+x)^{1/3}}{(1+x)^{2/3}(1-x)^{2/3}}\\ &=\frac{1}{2}\frac{(1+x)^{1/3}-(1-x)^{1/3}}{(1+x)^{2/3}(1-x)^{2/3}} + \frac{1}{2}\frac{(1+x)^{1/3}+(1-x)^{1/3}}{(1+x)^{2/3}(1-x)^{2/3}} \end{split} $$ For the first term, you let $$ u=(1+x)^{1/3}+(1-x)^{1/3} $$ which magically will give you $$ -\frac{3}{2}\int\frac{1}{u}\,du=-\frac{3}{2}\log u+C=-\frac{3}{2}\log \bigl((1+x)^{1/3}+(1-x)^{1/3}\bigr)+C $$ For the second term, you let $$ v=\frac{(1+x)^{1/3}}{(1-x)^{1/3}} $$ which equally magically will give you $$ \frac{3}{2}\int\frac{1}{v^2-v+1}\,du=\sqrt{3}\arctan\Bigl(\frac{2u-1}{\sqrt{3}}\Bigr)+C=\sqrt{3}\arctan\biggl(\frac{2\frac{(1+x)^{1/3}}{(1-x)^{1/3}}-1}{\sqrt{3}}\biggr)+C $$ What is left is to insert limits. I get $$ \frac{1}{2}\sqrt{3}\pi-\frac{1}{2}\log 2 $$ from $x=1$, and $$ \frac{1}{6}\sqrt{3}\pi+\frac{1}{2}\log 2 $$ from the negative of $x=-1$.

Adding them, we get

$$ \int_{-1}^1 \frac{1}{(1+x)^{1/3}(1-x)^{2/3}}\,dx=\frac{2}{\sqrt{3}}\pi $$

Answer 4

First, split $\displaystyle\int_{-1}^1$ into $\displaystyle\int_{-1}^0$ and $\displaystyle\int_0^1$. On the former, let $x\mapsto-x$. Then, on both, let $x=\cos2t$,

and employ the well-known trigonometric formulas for $1\pm\cos2t$. Now unite both integrals

again into a single one, using the fact that $\cot t$ for $0<t<\dfrac\pi4$ takes the same values as $\tan t$

for $\dfrac\pi4<t<\dfrac\pi2~.~$ Lastly, by way of a simple trigonometric substitution, rewrite the obtained

Wallis' integral in terms of the beta function, and use Euler's reflection formula for the $\Gamma$

function to arrive at the desired result.