Improper integral $ \int_a^b \frac{\sqrt{x-a}}{\sqrt{b-x}}dx $

Question

$$ \int_a^b \frac{\sqrt{x-a}}{\sqrt{b-x}}dx $$ I have this integral and I have no idea what substitution to make.I'm guessing that I could substitute u with the inverse of the function inside the integral,u=$ \frac{\sqrt{b-x}}{\sqrt{x-a}}$. Is there another way?

Answer 1

Hint - Put $x=a {\cos}^{2}\theta + b \sin^{2}\theta$ $$dx=2\sin\theta \cos\theta (b-a)d\theta$$ now after changing the limits $$\int_{a}^{b} \sqrt{\dfrac{x-a}{b-x}}dx=(b-a)\int_{0}^{\dfrac{\pi}{2}} \sqrt{\dfrac{a{\cos}^{2}\theta+b{\sin}^{2}\theta-a}{b-a{\cos}^{2}\theta -b{\sin}^{2}\theta}}(2\sin\theta \cos\theta)d\theta$$ $$\Rightarrow (b-a) \int_{0}^{\dfrac{\pi}{2}} \sqrt{\dfrac{(b-a){\sin}^{2}\theta}{(b-a){\cos}^{2}\theta}}(2\sin\theta \cos\theta)d\theta=2(b-a)\int_{0}^{\dfrac{\pi}{2}} {\sin}^{2}\theta d\theta=\dfrac{\pi (b-a)}{2}$$

please do tell me if my answer is correct. (I am new to calculus)

Answer 2

Substitute $u^2 = \frac{{x-a}}{{b-x}}$

$$ \begin{align} \int_a^b \frac{\sqrt{x-a}}{\sqrt{b-x}}dx &= (b-a)\int_0^{\infty} \frac{2u^2}{(1+u^2)^2}du \\ &\overset{ibp}=(b-a)\int_0^{\infty}\frac{1}{1+u^2} du =\frac{\pi}2(b-a)\\ \end{align}$$

Answer 3

Let $\sqrt{\frac{x-a}{{b-x}}}=\tan t$ and then $x=a\cos^2t+b\sin^2t$. So $$ dx=2(b-a)\sin t\cos tdt $$ and hence $$ \int_a^b\sqrt{\frac{x-a}{{b-x}}}dx=\int_0^{\pi/2}2(b-a)\sin t\cos t \tan tdt=2(b-a)\int_0^{\pi/2}\sin^2tdt=2(b-a)\frac{\pi}{4}=\dfrac{\pi (b-a)}{2}.$$

Answer 4

$$ \int_a^b \frac{\sqrt{x-a}}{\sqrt{b-x}}dx =\int_0^{b-a} \frac{\sqrt x}{\sqrt{b-a-x}}dx=(b-a)\int_0^1\frac{\sqrt x}{\sqrt{1-x}}dx $$ pulls the constants out.

Then with $1-x=t^2$,

$$\int_0^1\frac{\sqrt x}{\sqrt{1-x}}dx=2\int_0^1\sqrt{1-t^2}\,dt$$ which is half the area of a unit circle.