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Question Number 36538 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

∫(dx/((a+bx^2 )(√(b−ax^2 ))))

$$\int\frac{{dx}}{\left({a}+{bx}^{\mathrm{2}} \right)\sqrt{{b}−{ax}^{\mathrm{2}} \:}}\:\: \\ $$

Commented by abdo.msup.com last updated on 03/Jun/18

we have b −ax^2 =b(1−((√(a/b)) x)^2 if (a/b)>0 changement (√(a/b)) x =sin(t) give (a/b) x =sin^2 t ⇒x =(b/a) sin^2 t ⇒ I = ∫ (1/((a +b (b^2 /a^2 ) sin^4 t)(√(b.)) (√(a/b))cost))((2b)/a)sint costdt =((2b)/a) ∫ ((sint)/((√a)( a^3 +b^3 sin^4 t)))a^2 dt = ((2b)/(√a)) ∫ ((sint)/(a^3 +b^3 sin^4 t))dt...be continued...

$${we}\:{have}\:{b}\:−{ax}^{\mathrm{2}} ={b}\left(\mathrm{1}−\left(\sqrt{\frac{{a}}{{b}}}\:{x}\right)^{\mathrm{2}} \:\:{if}\:\frac{{a}}{{b}}>\mathrm{0}\right. \\ $$$${changement}\:\sqrt{\frac{{a}}{{b}}}\:{x}\:={sin}\left({t}\right)\:{give} \\ $$$$\:\frac{{a}}{{b}}\:{x}\:={sin}^{\mathrm{2}} {t}\:\Rightarrow{x}\:=\frac{{b}}{{a}}\:{sin}^{\mathrm{2}} {t}\:\Rightarrow \\ $$$${I}\:=\:\int\:\:\:\:\:\frac{\mathrm{1}}{\left({a}\:+{b}\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:{sin}^{\mathrm{4}} {t}\right)\sqrt{{b}.}\:\sqrt{\frac{{a}}{{b}}}{cost}}\frac{\mathrm{2}{b}}{{a}}{sint}\:{costdt} \\ $$$$=\frac{\mathrm{2}{b}}{{a}}\:\int\:\:\:\:\:\:\frac{{sint}}{\sqrt{{a}}\left(\:{a}^{\mathrm{3}} \:+{b}^{\mathrm{3}} {sin}^{\mathrm{4}} {t}\right)}{a}^{\mathrm{2}} {dt} \\ $$$$=\:\frac{\mathrm{2}{b}}{\sqrt{{a}}}\:\int\:\:\:\:\:\frac{{sint}}{{a}^{\mathrm{3}} \:+{b}^{\mathrm{3}} {sin}^{\mathrm{4}} {t}}{dt}...{be}\:{continued}... \\ $$

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