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We have to calculate value of the following integral : $$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) : \begin{align} & = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\ & = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\ & = \int_0^1\cfrac{dx}{\sqrt {(x^2-x+\frac 14)-\frac 14 }} \\ & = \int_0^1\cfrac{dx}{\sqrt {(x-\frac 12)^2-(\frac 12)^2 }} \\ & = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 12\sec \theta)^2-(\frac 12)^2 }} I\ used\ trigonometric\ substitution \ u=a\sec \theta, by \ it's \ form \ u^2-a^2 \\ & = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 14\sec^2 \theta)-\frac 14 }} \\ & = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 14(\sec ^2\theta-1)}} \ using \\tan^2\theta=\sec^2\theta-1 \\ & = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 12(\sqrt{\tan^2\theta) }}} \\ & = \int_0^1\sec\theta d\theta = \sec\theta \tan \theta |_0^1 \\ \end{align}

But i got problems calculating $\theta$ value, using trigonometric substitution, any help?

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    $\begingroup$ Sign error: $x - x^2 = 1/4 - (x-1/2)^2$. This invalidates the rest of the analysis. $\endgroup$ – eyeballfrog Apr 17 at 18:21
  • $\begingroup$ I noticed that, used $\sin$ substitution instead of $\sec$, due to the minus sign, and i get to the answer, $\pi$ thanks. $\endgroup$ – Sebastian Fernandez Apr 17 at 18:55
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    $\begingroup$ Just substitute $x\to \sin(t)^2$, then $x\in (0,1) \to t \in (0,\frac{\pi}{2})$, $dx \to 2 \sin(t) \cos(t)dt$ cancels against the denominator, and the result is $2 \frac{\pi}{2} =\pi$ $\endgroup$ – Dr. Wolfgang Hintze Apr 17 at 20:05
  • $\begingroup$ It is the beta function $B(\frac{1}{2},\frac{1}{2})=\pi$. $\endgroup$ – StubbornAtom May 27 at 21:05
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Are you sure that you got the correct antiderivative for the integral of the secant function? The correct indefinite integral of secant is $\int\sec{x}=\ln{|\tan{x}+\sec{x}|}+C$. Thus:

$$ \int_{0}^{1}\sec{\theta}\,d\theta=\ln{|\tan{\theta}+\sec{\theta}|}\bigg|_{0}^{1}. $$

EDIT:

Also note that:

$$ x-x^2=-(x^2-x)=-\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)=\\ -\left(\left[x-\frac{1}{2}\right]^2-\frac{1}{4}\right)= \frac{1}{4}-\left(x-\frac{1}{2}\right)^2. $$

So, I think the substitution that you should be using would be: $$ x=\frac{1}{2}\sin{\theta}+\frac{1}{2},\\ dx=\frac{1}{2}\cos{\theta}\,d\theta,\\ x=\frac{1}{2}\sin{\theta}+\frac{1}{2}\implies\theta=\arcsin{(2x-1)},\ -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}. $$

Putting it all together, you get the following:

$$ \int\frac{1}{\sqrt{x-x^2}}\,dx=\int\frac{1}{\sqrt{\frac{1}{4}-\left(\frac{1}{2}\sin{\theta}+\frac{1}{2}-\frac{1}{2}\right)^2}}\frac{1}{2}\cos{\theta}\,d\theta=\\ \int\frac{1}{\frac{1}{2}\sqrt{1-\sin^2{\theta}}}\frac{1}{2}\cos{\theta}\,d\theta= \int\frac{\cos{\theta}}{|\cos{\theta}|}\,d\theta= \int\frac{\cos{\theta}}{\cos{\theta}}\,d\theta=\\ \int\,d\theta=\theta+C= \arcsin{(2x-1)}+C.\\ \int_0^1\frac{1}{\sqrt {x(1-x)}}\,dx=\arcsin{(2x-1)}\bigg|_0^1=\\ \arcsin{(1)}-\arcsin{(-1)}=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi. $$

Wolfram Alpha check.

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You must convert the limits to be in terms of $\theta$. When $x=0,1$, what is $\theta$?

Also if you want another method consider partial fractions

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As $0\le x\le1$

WLOG $x=\sin^2t;0\le t\le\dfrac\pi2,dx=?$

So $\sqrt x=+\sin t,\sqrt{1-x}=?$

Alternatively $4x(1-x)=1-(2x-1)^2$

Set $2x-1=\sin y$ or $\cos y$

Observe that $2x-1=-\cos2t$ in the first method

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