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

2 Answers
May 18, 2017

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.

May 18, 2017

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.!