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Evaluate $\int \frac{dx}{\sqrt{\left ( x-a \right )\left ( b-x \right )}}$

Option 1)
$2\sin^{-1}\sqrt{\left ( \frac{x-a}{b-a} \right )}+c$

Option 2)
$2\cos^{-1}\sqrt{\left ( \frac{x-a}{b-a} \right )}+c$

Option 3)
$2\tan^{-1}\sqrt{\left ( \frac{x-a}{b-a} \right )}+c$

Option 4)
$2\tan^{-1}\sqrt{\left ( \frac{x-a}{b-a} \right )}+c$

Answers (1)

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As we learnt

Special type of indefinite integration -

Integrals of the form :

(i) $f(\sqrt{a^{2}-x^{2}})$ , (ii) $f(\sqrt{x^{2}-a^{2}})$

(iii) $f(\sqrt{a^{2}+x^{2}})$ , (iv) $f(a^{2}+x^{2})$

(v) $f\left ( \sqrt{\frac{a-x}{a+x}}\right )$ , (vi) $f\left ( \sqrt{\frac{a+x}{a-x}}\right )$

(vii) $f\left ( \sqrt{\frac{x-a}{b-x}} \right )$ , (viii) $f\left ( \sqrt{(x-a)(x-b)}\right )$

- wherein

Working rule :

for (i) put $x=a\sin \Theta$ or $a \cos \Theta$

for (ii) Put $x=a\sec \Theta$ or $a \, cosec\Theta$

for (iii) and (iv) Put $x=a\tan \Theta$ or $a\cot \Theta$

for (v) and (vi) Put $x=a\cos 2 \Theta$

for (vii) and (viii) Put $x=a\cos ^{2}\Theta +b\sin ^{2}\Theta$

Writing x = acos²q + bsin²q = a + (b - a) sin²q, the given integral becomes

$I\; = \;\int {\frac{{2(b - a)\;\sin \theta \cos \theta d\theta \,}}{{{{\left\{ {(a\;{{\cos }^2}\theta \; + \;b\,{{\sin }^2}\theta - a)\;\;\;(a\;{{\cos }^2}\theta \; + \;b\;{{\sin }^2}\theta \; - \;b} \right\}}^{1/2}}}}}$

$=\int {\frac{{2\left( {b - a} \right)\sin \theta \cos \theta d\theta }}{{\left( {b - a} \right)\sin \theta \cos \theta }}}$ $=\;\left( {\frac{{b - a}}{{b - a}}} \right)\;\;\int {2\;d\theta }$

$=2\theta +c=2{\sin ^{ - 1}}\sqrt {\left( {\frac{{x - a}}{{b - a}}} \right)} + c$

Option 1)

$2\sin^{-1}\sqrt{\left ( \frac{x-a}{b-a} \right )}+c$

Option 2)

$2\cos^{-1}\sqrt{\left ( \frac{x-a}{b-a} \right )}+c$

Option 3)

$2\tan^{-1}\sqrt{\left ( \frac{x-a}{b-a} \right )}+c$

Option 4)

$2\tan^{-1}\sqrt{\left ( \frac{x-a}{b-a} \right )}+c$

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