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Question Number 33120 by abdo imad last updated on 10/Apr/18

let give α>0 find the value of  ∫_0 ^1     (dx/(√((1−x)(1+αx)))) .

$${let}\:{give}\:\alpha>\mathrm{0}\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+\alpha{x}\right)}}\:. \\ $$

Answered by MJS last updated on 11/Apr/18

(1−x)(1+αx)=−αx^2 +(α−1)x+1  ∫(dx/(√(ax^2 +bx+c)))=−(1/(√(−a)))sin^(−1) ((2ax+b)/(√(b^2 −4ac))); a<0  ∫_0 ^1 (dx/(√(−αx^2 +(α−1)x+1)))=[−(1/(√α))sin^(−1) ((−2αx+α−1)/(√((α−1)^2 +4α)))]_0 ^1 =  =−(1/(√α))[sin^(−1) ((−2αx+α−1)/(α+1))]_0 ^1 =  =−(1/(√α))(sin^(−1) ((−α−1)/(α+1))−sin^(−1) ((α−1)/(α+1)))=  =(1/(√α))(sin^(−1) 1+sin^(−1) ((α−1)/(α+1)))=  =(1/(√α))((π/2)+sin^(−1) ((α−1)/(α+1)))

$$\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+\alpha{x}\right)=−\alpha{x}^{\mathrm{2}} +\left(\alpha−\mathrm{1}\right){x}+\mathrm{1} \\ $$ $$\int\frac{{dx}}{\sqrt{{ax}^{\mathrm{2}} +{bx}+{c}}}=−\frac{\mathrm{1}}{\sqrt{−{a}}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}{ax}+{b}}{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}};\:{a}<\mathrm{0} \\ $$ $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\sqrt{−\alpha{x}^{\mathrm{2}} +\left(\alpha−\mathrm{1}\right){x}+\mathrm{1}}}=\left[−\frac{\mathrm{1}}{\sqrt{\alpha}}\mathrm{sin}^{−\mathrm{1}} \frac{−\mathrm{2}\alpha{x}+\alpha−\mathrm{1}}{\sqrt{\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\alpha}}\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$ $$=−\frac{\mathrm{1}}{\sqrt{\alpha}}\left[\mathrm{sin}^{−\mathrm{1}} \frac{−\mathrm{2}\alpha{x}+\alpha−\mathrm{1}}{\alpha+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$ $$=−\frac{\mathrm{1}}{\sqrt{\alpha}}\left(\mathrm{sin}^{−\mathrm{1}} \frac{−\alpha−\mathrm{1}}{\alpha+\mathrm{1}}−\mathrm{sin}^{−\mathrm{1}} \frac{\alpha−\mathrm{1}}{\alpha+\mathrm{1}}\right)= \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\alpha}}\left(\mathrm{sin}^{−\mathrm{1}} \mathrm{1}+\mathrm{sin}^{−\mathrm{1}} \frac{\alpha−\mathrm{1}}{\alpha+\mathrm{1}}\right)= \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\alpha}}\left(\frac{\pi}{\mathrm{2}}+\mathrm{sin}^{−\mathrm{1}} \frac{\alpha−\mathrm{1}}{\alpha+\mathrm{1}}\right) \\ $$

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