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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}

d x = a \frac{2 t (t^{2} + 1) - 2 t (t^{2} - 1)}{{(t^{2} + 1)}^{2}} = a \frac{4 t}{{(t^{2} + 1)}^{2}}

I = \int (t + \frac{1}{t}) (a \frac{4 t}{{(t^{2} + 1)}^{2}}) = 4 a \int \frac{d t}{t^{2} + 1} =

= 4 {a t g}^{- 1} (t) + C = 4 {a t g}^{- 1} (\sqrt{\frac{a + x}{a - x}}) + 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 = t g θ \Rightarrow d t = (1 + {t g}^{2} θ) d θ

I = \int \frac{d t}{{(1 + t^{2})}^{2}} = \int \frac{(1 + {t g}^{2} θ) d θ}{{(1 + {t g}^{2} θ)}^{2}} = \int \frac{d θ}{1 + {t g}^{2} θ}

= \int {c o s}^{2} θ d θ = \int \frac{1 + c o s 2 θ}{2} d θ = \int (\frac{1}{2} + \frac{1}{2} c o s 2 θ) d θ =

= \frac{1}{2} θ + \frac{1}{4} s i n 2 θ + C = \frac{1}{2} θ + \frac{1}{2} s i n θ . c o s θ + C =

= \frac{1}{2} {t g}^{- 1} t + \frac{1}{2} \frac{t}{\sqrt{1 + t^{2}}} . \frac{1}{\sqrt{1 + t^{2}}} + C =

= \frac{1}{2} ({t g}^{- 1} t + \frac{t}{1 + t^{2}}) + C .