Question

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

Answer 1

Rearrange. Substitute. Voila!

Explanation:

Since `sin^2x + cos^2x = 1` , which is equivalent to `1-cos^2x = 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:

`x/5 = cost`

and hence

`x = 5cost`
`dx = -5sintdt`

Our integral with the variable t now reads:

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

Whence we rearrange and get:

`int (-sintdt)/(5costsint)`,

which is:

`-1/5 int (dt)/(cost)`

And this integral is solved in the following way:

We make a small but cunning rearrangement, namely:

`1/cost = 1/cost * 1 = 1/cost * cost/cost = cost/cos^2t = cost/(1-sin^2t)`,

So that our integral now reads :

`-1/5 int (costdt)/(1-sin^2t)`

Why this? Because if we now set `sint = u` then `du = costdt` and with this we have:

`-1/5 int (du)/(1-u^2)`

Which is just :

`-1/5 1/2 (int (du)/(1+u) + int (du)/(1-u))` ,

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

`-1/5 1/2 ( ln(1+u) -ln(1-u))`

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

Answer 2

`1/5ln|(5-sqrt(25-x^2))/x|+C.`

Explanation:

Let us subst. `x=5sint rArr dx=5cost dt.`

`:. I=int1/(xsqrt(25-x^2))dx`

`=int1/(5sintsqrt(25-25sin^2t)) 5costdt,`

`=intcost/{(sint)(5cost)}dt,`

`=1/5intcsctdt,`

`=1/5ln|csct-cott|,`

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

`I=1/5ln|(5-sqrt(25-x^2))/x|+C.`

Enjoy Maths.!