# int sqrt((a+x)/(a-x))dx?

Jun 6, 2017

x + c

#### Explanation:

$\int \sqrt{\frac{a + x}{a - x}} \mathrm{dx}$ is to be integrated ,so lets rationalize it.

int sqrt((a+x)/(a-x) x $\sqrt{\frac{a - x}{a + x}} \mathrm{dx}$

=$\int \frac{\sqrt{{a}^{2} - {x}^{2}}}{\sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx}$=$\int 1 \mathrm{dx}$= x+c

Jun 6, 2017

$a \cdot a r c \sin \left(\frac{x}{a}\right) - \sqrt{{a}^{2} - {x}^{2}} + C .$

#### Explanation:

Let, $I = \int \sqrt{\frac{a + x}{a - x}} \mathrm{dx} .$

Rationalising, we have,

$I = \int \sqrt{\frac{a + x}{a - x}} \times \sqrt{\frac{a + x}{a + x}} \mathrm{dx}$

$= \int \frac{a + x}{\sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx}$

$= a \int \frac{1}{\sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx} - \frac{1}{2} \int \frac{- 2 x}{\sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx}$

$= a \cdot a r c \sin \left(\frac{x}{a}\right) - \frac{1}{2} \int \left\{\frac{d}{\mathrm{dx}} \left({a}^{2} - {x}^{2}\right) \cdot {\left({a}^{2} - {x}^{2}\right)}^{- \frac{1}{2}}\right\} \mathrm{dx}$

$= a \cdot a r c \sin \left(\frac{x}{a}\right) - \frac{1}{2} {\left({a}^{2} - {x}^{2}\right)}^{- \frac{1}{2} + 1} / \left(- \frac{1}{2} + 1\right)$

$= a \cdot a r c \sin \left(\frac{x}{a}\right) - {\left({a}^{2} - {x}^{2}\right)}^{\frac{1}{2}}$

$\Rightarrow I = a \cdot a r c \sin \left(\frac{x}{a}\right) - \sqrt{{a}^{2} - {x}^{2}} + C .$

Note that, the later integral has been derived using the following

useful Result

Result : $\int {\left[f \left(x\right)\right]}^{n} \cdot f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c , n \ne - 1.$

Enjoy Maths.!

Jun 8, 2017

$- a \cdot a r c \cos \left(\frac{x}{a}\right) - \sqrt{{a}^{2} - {x}^{2}} + C .$

#### Explanation:

As an Aliter, let us solve this using a Trgo. Substn.

$x = a \cos 2 y \Rightarrow \mathrm{dx} = - 2 a \sin 2 y \mathrm{dy} .$

Also, $\sqrt{\frac{a + x}{a - x}} = \sqrt{\frac{a + a \cos 2 y}{a - a \cos 2 y}} = \cot y .$

$\therefore I = \int \left(\cot y\right) \left(- 2 a \sin 2 y\right) \mathrm{dy} .$

$= - 2 a \int 2 {\cos}^{2} y \mathrm{dy}$

$= - 2 a \int \left(1 + \cos 2 y\right) \mathrm{dy}$

$= - 2 a \left(y + \sin \frac{2 y}{2}\right)$

$= - a \left(2 y\right) - a \sin 2 y$

$= - a \left(2 y\right) - \sqrt{{a}^{2} - {a}^{2} {\cos}^{2} 2 y}$

$\Rightarrow I = - a \cdot a r c \cos \left(\frac{x}{a}\right) - \sqrt{{a}^{2} - {x}^{2}} + C .$

Enjoy Maths.!