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Question Number 12130 by sujith last updated on 14/Apr/17

∫(√(a+x/a−x )) −(√(a−x/a+x))

$$\int\sqrt{{a}+{x}/{a}−{x}\:} \\ $$$$−\sqrt{{a}−{x}/{a}+{x}} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

((a+x)/(a−x))=t^2 ⇒((a+x−a+x)/(a+x+a−x))=((t^2 −1)/(t^2 +1))⇒x=a((t^2 −1)/(t^2 +1)) dx=a((2t(t^2 +1)−2t(t^2 −1))/((t^2 +1)^2 ))=a((4t)/((t^2 +1)^2 )) I=∫(t+(1/t))(a((4t)/((t^2 +1)^2 )))=4a∫(dt/(t^2 +1))= =4atg^(−1) (t)+C=4atg^(−1) ((√((a+x)/(a−x))))+C

$$\frac{{a}+{x}}{{a}−{x}}={t}^{\mathrm{2}} \Rightarrow\frac{{a}+{x}−{a}+{x}}{{a}+{x}+{a}−{x}}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\Rightarrow{x}={a}\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${dx}={a}\frac{\mathrm{2}{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{2}{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }={a}\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${I}=\int\left({t}+\frac{\mathrm{1}}{{t}}\right)\left({a}\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right)=\mathrm{4}{a}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\mathrm{4}{atg}^{−\mathrm{1}} \left({t}\right)+{C}=\mathrm{4}{atg}^{−\mathrm{1}} \left(\sqrt{\frac{{a}+{x}}{{a}−{x}}}\right)+{C} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

t=tgθ⇒dt=(1+tg^2 θ)dθ I=∫(dt/((1+t^2 )^2 ))=∫(((1+tg^2 θ)dθ)/((1+tg^2 θ)^2 ))=∫(dθ/(1+tg^2 θ)) =∫cos^2 θdθ=∫((1+cos2θ)/2)dθ=∫((1/2)+(1/2)cos2θ)dθ= =(1/2)θ+(1/4)sin2θ+C=(1/2)θ+(1/2)sinθ.cosθ+C= =(1/2)tg^(−1) t+(1/2)(t/(√(1+t^2 ))).(1/(√(1+t^2 )))+C = =(1/2)(tg^(−1) t+(t/(1+t^2 )))+C .■

$${t}={tg}\theta\Rightarrow{dt}=\left(\mathrm{1}+{tg}^{\mathrm{2}} \theta\right){d}\theta \\ $$$${I}=\int\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }=\int\frac{\left(\mathrm{1}+{tg}^{\mathrm{2}} \theta\right){d}\theta}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }=\int\frac{{d}\theta}{\mathrm{1}+{tg}^{\mathrm{2}} \theta} \\ $$$$=\int{cos}^{\mathrm{2}} \theta{d}\theta=\int\frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{2}}{d}\theta=\int\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{2}\theta\right){d}\theta= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\theta+\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{2}\theta+{C}=\frac{\mathrm{1}}{\mathrm{2}}\theta+\frac{\mathrm{1}}{\mathrm{2}}{sin}\theta.{cos}\theta+{C}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{tg}^{−\mathrm{1}} {t}+\frac{\mathrm{1}}{\mathrm{2}}\frac{{t}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}+{C}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({tg}^{−\mathrm{1}} {t}+\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)+{C}\:\:.\blacksquare \\ $$

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