Integral Calculator: $$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}$$

[Also sprach Zarathustra]

Calculate the following integral:

$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$

 

[Sherlock1]

Calculate the following integral:

$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$

Let $\displaystyle x = a~\sin^2{t}+b~\cos^2{t} \implies \begin{cases} x = (a-b)\sin^2{t}+b
\\ x = a+(b-a)\cos^2{t}
\\ dx = 2(a-b)\sin{t}\cos{t}
\end{cases}$ and $t= \cos^{-1}\left(\frac{\sqrt{x-a}}{\sqrt{b-a}}\right)$.Thus $\displaystyle I = \int_{0}^{\pi/2}\frac{2(b-a)\sin{t}\cos{t}}{\sqrt{b-a}\sqrt{b-a}\sin{t}\cos{t}}\;{dt} = \frac{2(b-a)}{(b-a)}\int_{0}^{\pi/2}\;{dt} = 2 \left(\frac{\pi}{2}\right) = \pi.$

 

[Prove It]

Calculate the following integral:

$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$

\[ \displaystyle \begin{align*} \int_a^b{\frac{dx}{\sqrt{(x - a)(b - x)}}} &= \int_a^b{\frac{dx}{\sqrt{-x^2 + (a + b)x - ab}}} \\ &= \int_a^b{\frac{dx}{\sqrt{-\left[ x^2 - (a + b)x + ab \right]}}} \\ &= \int{\frac{dx}{\sqrt{-\left\{ x^2 - (a + b)x + \left[-\left(\frac{a + b}{2}\right)\right]^2 - \left[-\left(\frac{a + b}{2}\right)\right]^2 + ab \right\} }}} \\ &= \int_a^b{\frac{dx}{\sqrt{-\left\{ \left[ x - \left( \frac{a+b}{2} \right) \right]^2 - \frac{(a + b)^2}{4} + \frac{4ab}{4} \right\}}}} \\ &= \int_a^b{\frac{dx}{\sqrt{-\left\{ \left[ x - \left( \frac{a + b}{2} \right) \right]^2 + \frac{4ab - a^2 - 2ab - b^2}{4} \right\} }}} \\ &= \int_a^b{\frac{dx}{\sqrt{- \left\{ \left[ x - \left(\frac{a + b}{2}\right) \right]^2 - \frac{a^2 - 2ab + b^2}{4} \right\} }}} \\ &= \int_a^b{ \frac{dx}{\sqrt{ - \left\{ \left[ x - \left(\frac{a + b}{2}\right) \right]^2 - \left(\frac{a - b}{2}\right)^2 \right\} }} } \\ &= \int_a^b{\frac{dx}{\sqrt{ \left( \frac{a - b}{2} \right)^2 - \left[ x - \left(\frac{a + b}{2}\right) \right]^2 }}} \end{align*} \]

Now make the substitution $ x - \left(\frac{a + b}{2}\right) = \left(\frac{a - b}{2}\right)\sin{\theta} \implies dx = \left(\frac{a - b}{2}\right)\cos{\theta}\,d\theta $, and note that when $x = a, \theta = \frac{\pi}{2}$ and when $x = b, \theta = \frac{3\pi}{2}$ and the integral becomes

\[ \displaystyle \begin{align*} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{ \frac{\left(\frac{a - b}{2}\right)\cos{\theta}\,d\theta}{\sqrt{ \left(\frac{a - b}{2}\right)^2 - \left[\left(\frac{a - b}{2}\right)\sin{\theta}\right]^2} } } &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\frac{\left(\frac{a - b}{2}\right)\cos{\theta}\,d\theta}{\sqrt{ \left( \frac{a - b}{2} \right)^2 - \left( \frac{a - b}{2} \right)^2 \sin^2{\theta} } } } \\ &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\frac{\left(\frac{a - b}{2}\right)\cos{\theta}\,d\theta}{\sqrt{ \left(\frac{a - b}{2}\right)^2\left(1 - \sin^2{\theta} \right) }}} \\ &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\frac{\left(\frac{a - b}{2}\right)\cos{\theta}\,d\theta}{\sqrt{ \left(\frac{a - b}{2}\right)^2\cos^2{\theta} }}} \\ &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\frac{\left(\frac{a - b}{2}\right)\cos{\theta}\,d\theta}{\left(\frac{a - b}{2}\right)\cos{\theta}}} \\ &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{1\,d\theta} \\ &= \left[\theta\right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \\ &= \frac{3\pi}{2} - \frac{\pi}{2} \\ &= \pi \end{align*} \]

Q.E.D. :)

 

[Markov2]

The most important thing in order to make this work: $b>a.$

 

[Prove It]

The most important thing in order to make this work: $b>a.$

Yes, that's what I had assumed, but you're right, it's important to note that :)

 

[oasi]

The most important thing in order to make this work: $b>a.$

if we miss this point ,we are going to do big mistake

 

[alyafey22]

Calculate the following integral:

$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$

That might seems a bit strange but this can be solved using the beta function !

I will make a substitution first ... $$\text{let : }x = a+(b-a) t$$

$$\int^1_0 \frac{(b-a) \, dt}{\sqrt{(a+(b-a) t - a) (b-a-(b-a) t)}}$$

$$\int^1_0 \frac{(b-a) \, dt}{(b-a) \sqrt{t(1- t)}}$$

$$\int^1_0 \frac{(b-a) \, dt}{(b-a) \sqrt{t(1- t)}}$$

$$\int^1_0 \, t^{-\frac{1}{2}}(1-t)^{-\frac{1}{2}}\, dt\, = \, \Gamma(\frac{1}{2})\Gamma(\frac{1}{2})= \sqrt{\pi}\cdot \sqrt{\pi}= \pi$$