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Question Number 122697 by john santu last updated on 19/Nov/20

$$\int \frac{dx}{\sqrt{(x-a)(b-x)}}?$$

Answered by liberty last updated on 19/Nov/20

$$Solve\phi \left(x\right)=\int \frac{dx}{\sqrt{(x-a)(b-x)}}?$$$$Solution:$$$$lettingx=a\cos {}^{2}q+b\sin {}^{2}q$$$$wherea<x<b,0<q<\frac{\pi }{2}$$$$weget\{\begin{array}{l}x-a=(b-a)\sin {}^{2}q\\ b-x=(b-a)\cos {}^{2}q\end{array}$$$$and\tan {}^{2}q=\frac{x-a}{b-x}or\tan q=\sqrt{\frac{x-a}{b-x}}$$$$dx=2(b-a)\sin q\cos qdq$$$$implyingthat$$$$\phi \left(x\right)=\int \frac{dx}{\sqrt{(x-a)(b-x)}}=\int 2dq$$$$\phi \left(x\right)=2q+c=2\arctan \left(\sqrt{\frac{x-a}{b-x}}\right)+c.✓$$

Commented by john santu last updated on 19/Nov/20

$$nice$$$$$$

Answered by som(math1967) last updated on 19/Nov/20

$$\mathrm{let}\mathrm{x}-\mathrm{a}={\mathrm{z}}^2$$$$\mathrm{dx}=2\mathrm{zdz}$$$$\int \frac{2\mathrm{zdz}}{\mathrm{z}\sqrt{\mathrm{b}-\mathrm{a}-{\mathrm{z}}^2}}$$$$2\int \frac{\mathrm{dz}}{\sqrt{\mathrm{b}-\mathrm{a}-{\mathrm{z}}^2}}$$$${2\sin }^{-1}\frac{\mathrm{z}}{\sqrt{\mathrm{b}-\mathrm{a}}}+\mathrm{c}$$$${2\sin }^{-1}\left(\sqrt{\frac{\mathrm{x}-\mathrm{a}}{\mathrm{b}-\mathrm{a}}}\right)+\mathrm{c}$$

Answered by MJS_new last updated on 19/Nov/20

$$\int \frac{dx}{\sqrt{(x-a)(b-x)}}=$$$$[t=\frac{\sqrt{x-a}}{\sqrt{b-x}}\to dx=\frac{2}{b-a}\sqrt{x-a}\sqrt{{(b-x)}^3}dt]$$$$=2\int \frac{dt}{t^2+1}=2\arctan t=2\arctan \frac{\sqrt{x-a}}{\sqrt{b-x}}+C$$

Commented by liberty last updated on 19/Nov/20

$$amazing...yourmethodveryfastest$$$$sir$$

Commented by MJS_new last updated on 19/Nov/20

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{just}\mathrm{an}\mathrm{old}\mathrm{integral}\mathrm{lover}...$$

Commented by liberty last updated on 19/Nov/20

$$hahahaha....great$$

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