# How can you proof int dx/(a^2-x^2)=1/(2a) log |(a+x)/(a-x)| + c using x = a sintheta ?

## $\int \frac{\mathrm{dx}}{{a}^{2} - {x}^{2}} = \frac{1}{2 a} \log | \frac{a + x}{a - x} | + c$ using $x = a \sin \theta$

Jul 19, 2018

Kindly refer to Explanation.

#### Explanation:

Suppose that, $I = \int \frac{1}{{a}^{2} - {x}^{2}} \mathrm{dx}$.

Using the substn., $x = a \sin \theta , s o , \mathrm{dx} = a \cos \theta d \theta$,

$I = \int \frac{a \cos \theta}{{a}^{2} - {a}^{2} {\sin}^{2} \theta} d \theta$,

$= \int \frac{a \cos \theta}{{a}^{2} {\cos}^{2} \theta} d \theta$,

$= \frac{1}{a} \int \frac{1}{\cos} \theta d \theta$,

$= \frac{1}{a} \int \sec \theta d \theta$,

$= \frac{1}{a} \ln | \left(\sec \theta + \tan \theta\right) |$,

$= \frac{1}{a} \ln | \left(\frac{1}{\cos} \theta + \sin \frac{\theta}{\cos} \theta\right) |$,

$= \frac{1}{a} \ln | \frac{1 + \sin \theta}{\cos} \theta |$,

$= \frac{1}{2 a} \left\{2 \ln | \frac{1 + \sin \theta}{\cos} \theta |\right\}$,

$= \frac{1}{2 a} \left\{\ln | \frac{1 + \sin \theta}{\cos} \theta {|}^{2}\right\}$,

$= \frac{1}{2 a} \ln | {\left(1 + \sin \theta\right)}^{2} / {\cos}^{2} \theta |$,

$= \frac{1}{2 a} \ln | {\left(1 + \sin \theta\right)}^{2} / \left(1 - {\sin}^{2} \theta\right) |$,

$= \frac{1}{2 a} \ln | {\left(1 + \sin \theta\right)}^{2} / \left(\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)\right) |$,

$= \frac{1}{2 a} \ln | \frac{1 + \sin \theta}{1 - \sin \theta} |$,

$= \frac{1}{2 a} \ln | \frac{a + a \sin \theta}{a - a \sin \theta} |$,

$= \frac{1}{2 a} \ln | \frac{a + x}{a - x} | + c$, as desired!

Jul 19, 2018

#### Explanation:

Normally, you perform this integral by the composition into partial fractions.

But you need.

$x = a \sin \theta$, $\implies$, $\mathrm{dx} = a \cos \theta d \theta$

${a}^{2} - {x}^{2} = {a}^{2} - {a}^{2} {\sin}^{2} \theta = {a}^{2} {\cos}^{2} \theta$

Therefore,

The integral is

$I = \int \frac{\mathrm{dx}}{{a}^{2} - {x}^{2}} = \int \frac{a \cos \theta d \theta}{{a}^{2} {\cos}^{2} \theta} = \frac{1}{a} \int \sec \theta d \theta$

$= \frac{1}{a} \int \frac{\sec \theta \left(\tan \theta + \sec \theta\right) d \theta}{\tan \theta + \sec \theta}$

$= \frac{1}{a} \int \frac{\left(\sec \theta \tan \theta + {\sec}^{2} \theta\right) d \theta}{\tan \theta + \sec \theta}$

Let $u = \tan \theta + \sec \theta$

$\implies$, $\mathrm{du} = \left(\sec \theta \tan \theta + {\sec}^{2} \theta\right) d \theta$

Therefore,

$I = \frac{1}{a} \int \frac{\mathrm{du}}{u}$

$= \frac{1}{a} \ln \left(u\right)$

$= \frac{1}{a} \ln \left(\tan \theta + \sec \theta\right)$

$= \frac{1}{a} \ln \left(\frac{a}{\sqrt{{a}^{2} - {x}^{2}}} + \frac{x}{\sqrt{{a}^{2} - {x}^{2}}}\right)$

$= \frac{1}{a} \ln \left(a + x\right) - \frac{1}{a} \ln \left(\sqrt{{a}^{2} - {x}^{2}}\right)$

$= \frac{1}{a} \ln \left(a + x\right) - \frac{1}{2 a} \ln \left({a}^{2} - {x}^{2}\right)$

$= \frac{2}{2 a} \ln \left(a + x\right) - \frac{1}{2 a} \ln \left({a}^{2} - {x}^{2}\right)$

$= \frac{1}{2} a \left(\ln \left({\left(a + x\right)}^{2} / \left({a}^{2} - {x}^{2}\right)\right)\right)$

$= \frac{1}{2} a \ln \left(\frac{\left(a + x\right) \left(a + x\right)}{\left(a + x\right) \left(a - x\right)}\right)$

$= \frac{1}{2 a} \ln | \frac{a + x}{a - x} | + C$

$Q E D$

Proof is below

#### Explanation:

Let $x = a \setminus \sin \setminus \theta \setminus \implies \mathrm{dx} = a \setminus \cos \setminus \theta \setminus d \setminus \theta$

$\setminus \therefore \setminus \int \setminus \frac{\mathrm{dx}}{{a}^{2} - {x}^{2}}$

$= \setminus \int \setminus \frac{a \setminus \cos \setminus \theta \setminus d \setminus \theta}{{a}^{2} - {a}^{2} \setminus {\sin}^{2} \setminus \theta}$

$= \setminus \int \setminus \frac{a \setminus \cos \setminus \theta \setminus d \setminus \theta}{{a}^{2} \setminus {\cos}^{2} \setminus \theta}$

$= \frac{1}{a} \setminus \int \setminus \frac{\setminus d \setminus \theta}{\setminus \cos \setminus \theta}$

$= \frac{1}{a} \setminus \int \setminus \sec \setminus \theta \setminus d \setminus \theta$

$= \frac{1}{a} \setminus \int \frac{\setminus \sec \setminus \theta \left(\setminus \sec \setminus \theta + \setminus \tan \setminus \theta\right)}{\setminus \sec \setminus \theta + \setminus \tan \setminus \theta} \setminus d \setminus \theta$

$= \frac{1}{a} \setminus \int \frac{\left(\setminus \sec \setminus \theta \setminus \tan \setminus \theta + \setminus {\sec}^{2} \setminus \theta\right) \setminus d \setminus \theta}{\setminus \sec \setminus \theta + \setminus \tan \setminus \theta}$

$= \frac{1}{a} \setminus \int \frac{d \left(\setminus \sec \setminus \theta + \tan \setminus \theta\right)}{\setminus \sec \setminus \theta + \setminus \tan \setminus \theta}$

$= \frac{1}{a} \setminus \ln | \setminus \sec \setminus \theta + \setminus \tan \setminus \theta | + c$

$= \frac{1}{a} \setminus \ln | \frac{1}{\setminus \cos \setminus \theta} + \frac{\setminus \sin \setminus \theta}{\setminus \cos \setminus \theta} | + c$

$= \frac{1}{a} \setminus \ln | \frac{1 + \setminus \sin \setminus \theta}{\setminus \cos \setminus \theta} | + c$

$= \frac{1}{a} \setminus \ln | \frac{1 + \frac{x}{a}}{\setminus \sqrt{1 - \setminus {\sin}^{2} \setminus \theta}} | + c$

$= \frac{1}{a} \setminus \ln | \frac{\frac{a + x}{a}}{\setminus \sqrt{1 - {x}^{2} / {a}^{2}}} | + c$

$= \frac{1}{a} \setminus \ln | \frac{a + x}{{a}^{2} - {x}^{2}} ^ \left\{\frac{1}{2}\right\} | + c$

$= \frac{1}{a} \setminus \cdot \frac{1}{2} \setminus \ln | \frac{{\left(a + x\right)}^{2}}{{a}^{2} - {x}^{2}} | + c$

$= \frac{1}{2 a} \setminus \ln | \frac{{\left(a + x\right)}^{2}}{\left(a + x\right) \left(a - x\right)} | + c$

$= \frac{1}{2 a} \setminus \ln | \frac{a + x}{a - x} | + c$