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Question Number 5878 by gourav~ last updated on 03/Jun/16

\int \sqrt{\frac{a - x}{a + x}} d x

p l e a s e h e l p \dots \dots . =

Answered by Yozzii last updated on 03/Jun/16

L e t I = \int \sqrt{\frac{a - x}{a + x}} d x .

\frac{a - x}{a + x} = \frac{a^{2} - x^{2}}{a^{2} + x^{2} + 2 a x} = j

L e t x = a c o s u \Rightarrow \frac{a^{2} - x^{2}}{{(a + x)}^{2}} = \frac{a^{2} (1 - {c o s}^{2} u)}{a^{2} {(1 + c o s u)}^{2}} (a > 0)

j = \frac{{s i n}^{2} u}{{(1 + c o s u)}^{2}} \Rightarrow \sqrt{j} = \sqrt{\frac{a - x}{a + x}} = \sqrt{\frac{{s i n}^{2} u}{{(1 + c o s u)}^{2}}}

I f I \in R \Rightarrow \frac{a - x}{a + x} ⩾ 0 \Rightarrow \frac{x - a}{x + a} ⩽ 0 \Rightarrow - a ⩽ x ⩽ a \Rightarrow - 1 ⩽ c o s u ⩽ 1

\Rightarrow 0 ⩽ c o s u + 1 ⩽ 2 ∴ \sqrt{{(1 + c o s u)}^{2}} = 1 + c o s u

∴ 0 ⩽ {c o s}^{2} u ⩽ 1

- 1 ⩽ - {c o s}^{2} u ⩽ 0

0 ⩽ 1 - {c o s}^{2} u ⩽ 1

\Rightarrow 0 ⩽ \sqrt{1 - {c o s}^{2} u} ⩽ 1 \Rightarrow 0 ⩽ s i n u ⩽ 1

∴ \sqrt{{s i n}^{2} u} = s i n u .

I n a l l, \sqrt{j} = \frac{s i n u}{1 + c o s u} .

x = a c o s u \Rightarrow d x = - a s i n u d u .

∴ I = \int \frac{- {a s i n}^{2} u}{1 + c o s u} d u .

I = \frac{- a}{2} \int \frac{4 {s i n}^{2} 0.5 {u c o s}^{2} 0.5 u}{{c o s}^{2} 0.5 u} d u (a l t e r n a t i v e l y, \frac{{s i n}^{2} u}{1 + c o s u} = \frac{{s i n}^{2} u (1 - c o s u)}{(1 + c o s u) (1 - c o s u)} = \frac{{s i n}^{2} u (1 - c o s u)}{1 - {c o s}^{2} u} = \frac{{s i n}^{2} u (1 - c o s u)}{{s i n}^{2} u} = 1 - c o s u)

I = - 2 a \int {s i n}^{2} 0.5 u d u

I = - 2 a \int 0.5 (1 - c o s u) d u

I = - a \int (1 - c o s u) d u

I = - a (u - s i n u) + C

I = a (\frac{\sqrt{a^{2} - x^{2}}}{a} - {c o s}^{- 1} \frac{x}{a}) + C

I = \sqrt{a^{2} - x^{2}} - {a c o s}^{- 1} \frac{x}{a} + C

Commented by Yozzii last updated on 03/Jun/16

A l t e r n a t i v e l y,

L e t \sqrt{x + 2} = 2 s i n u

\Rightarrow \frac{1}{2 \sqrt{x + 2}} d x = 2 c o s u d u

d x = 8 s i n u c o s u d u

x + 2 = 4 {s i n}^{2} u

x = 4 {s i n}^{2} u - 2

- x = 2 - 4 {s i n}^{2} u

2 - x = 4 - 4 {s i n}^{2} u = 4 {c o s}^{2} u \Rightarrow c o s u = \frac{\sqrt{2 - x}}{2}

I = \int \frac{2 c o s u}{2 s i n u} \times 8 s i n u c o s u d u

I = 8 \int {c o s}^{2} u d u

I = 4 \int (1 + c o s 2 u) d u

I = 4 (u + \frac{1}{2} s i n 2 u) + C

I = 4 u + 4 s i n u c o s u + C

I = 4 {s i n}^{- 1} \frac{\sqrt{x + 2}}{2} + \sqrt{4 - x^{2}} + C

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