# How do you integrate int 1/(xsqrt(25-x^2)) by trigonometric substitution?

May 18, 2017

Rearrange. Substitute. Voila!

#### Explanation:

Since ${\sin}^{2} x + {\cos}^{2} x = 1$ , which is equivalent to $1 - {\cos}^{2} x = {\sin}^{x}$, we want something like this in the denominator . For that we rewrite this into:

int dx/(x*5(sqrt(1-(x/5)^2))

Then we set:

$\frac{x}{5} = \cos t$

and hence

$x = 5 \cos t$
$\mathrm{dx} = - 5 \sin t \mathrm{dt}$

Our integral with the variable t now reads:

int (-5sintdt)/(25cost(sqrt(1-cost^2))

Whence we rearrange and get:

$\int \frac{- \sin t \mathrm{dt}}{5 \cos t \sin t}$,

which is:

$- \frac{1}{5} \int \frac{\mathrm{dt}}{\cos t}$

And this integral is solved in the following way:

We make a small but cunning rearrangement, namely:

$\frac{1}{\cos} t = \frac{1}{\cos} t \cdot 1 = \frac{1}{\cos} t \cdot \cos \frac{t}{\cos} t = \cos \frac{t}{\cos} ^ 2 t = \cos \frac{t}{1 - {\sin}^{2} t}$,

So that our integral now reads :

$- \frac{1}{5} \int \frac{\cos t \mathrm{dt}}{1 - {\sin}^{2} t}$

Why this? Because if we now set $\sin t = u$ then $\mathrm{du} = \cos t \mathrm{dt}$ and with this we have:

$- \frac{1}{5} \int \frac{\mathrm{du}}{1 - {u}^{2}}$

Which is just :

$- \frac{1}{5} \frac{1}{2} \left(\int \frac{\mathrm{du}}{1 + u} + \int \frac{\mathrm{du}}{1 - u}\right)$ ,

The one half comes about from the partial fraction decomposition. This is just :

$- \frac{1}{5} \frac{1}{2} \left(\ln \left(1 + u\right) - \ln \left(1 - u\right)\right)$

However this is a solution expressed through the variable u, but back substitution is done trivially if needed.

May 18, 2017

$\frac{1}{5} \ln | \frac{5 - \sqrt{25 - {x}^{2}}}{x} | + C .$

#### Explanation:

Let us subst. $x = 5 \sin t \Rightarrow \mathrm{dx} = 5 \cos t \mathrm{dt} .$

$\therefore I = \int \frac{1}{x \sqrt{25 - {x}^{2}}} \mathrm{dx}$

$= \int \frac{1}{5 \sin t \sqrt{25 - 25 {\sin}^{2} t}} 5 \cos t \mathrm{dt} ,$

$= \int \cos \frac{t}{\left(\sin t\right) \left(5 \cos t\right)} \mathrm{dt} ,$

$= \frac{1}{5} \int \csc t \mathrm{dt} ,$

$= \frac{1}{5} \ln | \csc t - \cot t | ,$

Since, sint=x/5, csc t=5/x, &, cott=sqrt(1-(x/5)^2)/(x/5),we have,

$I = \frac{1}{5} \ln | \frac{5 - \sqrt{25 - {x}^{2}}}{x} | + C .$

Enjoy Maths.!