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

Question

William Abdalla · Accepted Answer

Rearrange. Substitute. Voila!

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.

Samantha Adkison · Answer

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

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.!

Charles Arocha · Answer

undefined To integrate $ \int \frac{1}{x\sqrt{25-x^2}} $ by trigonometric substitution, use the substitution $ x = 5\sin(\theta) $. Then, $ dx = 5\cos(\theta) d\theta $. Substitute these into the integral, simplify, and integrate using trigonometric identities.