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

$$\int \sqrt{a+x/a-x}$$$$-\sqrt{a-x/a+x}$$

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

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

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

$$t=tg\theta \Rightarrow dt=(1+{tg}^2\theta )d\theta$$$$I=\int \frac{dt}{{(1+t^2)}^2}=\int \frac{(1+{tg}^2\theta )d\theta }{{(1+{tg}^2\theta )}^2}=\int \frac{d\theta }{1+{tg}^2\theta }$$$$=\int {cos}^2\theta d\theta =\int \frac{1+cos2\theta }{2}d\theta =\int (\frac{1}{2}+\frac{1}{2}cos2\theta )d\theta =$$$$=\frac{1}{2}\theta +\frac{1}{4}sin2\theta +C=\frac{1}{2}\theta +\frac{1}{2}sin\theta .cos\theta +C=$$$$=\frac{1}{2}{tg}^{-1}t+\frac{1}{2}\frac{t}{\sqrt{1+t^2}}.\frac{1}{\sqrt{1+t^2}}+C=$$$$=\frac{1}{2}({tg}^{-1}t+\frac{t}{1+t^2})+C.◼$$

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