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

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.

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.!