#" "#
I am assuming that your question is:
#color(red)(int (x^2)/[25-x^2]^(3/2) " "dx#
Rewrite the problem as follows:
#color(red)(int (x^2)/[[sqrt[25-x^2]]^(3) " "dx# #color(blue)(Exp.1#
Use Trigonometric SIN substitution
Let #color(green)(sqrt(a^2-x^2) rArr x = a sin theta#
Consider the denominator part [DR] of our problem:
#[sqrt[25-x^2]]^(3)#
#rArr [sqrt[[5^2-x^2]]]^3#
Using sin substitution, substitute #color(red)(x=5 sin theta# to obtain
#rArr [sqrt[[25-(5 sin theta)^2]]]^3#
#rArr [sqrt[[25-(5^2 sin^2 theta)]]]^3#
#rArr [sqrt[[25-25 sin^2 theta]]]^3#
#rArr [sqrt[[25(1-sin^2 theta)]]^3#
We use the following trigonometric relationship to simplify further:
#color(green)(sin^2 theta + cos^2 theta = 1#
#color(green)(cos^2 theta = 1 - sin^2 theta)#
#rArr [sqrt[25(cos^2 theta)]]^3#
The above expression reduces to
#color(red)([5 |cos theta|]^3)#
We consider #color(blue)(cos theta)# is positive
#rArr [ 5 cos theta ]^3#
#rArr 5^3 cos^3 theta#
#rArr 125 cos^3 theta#
We already have
#color(blue)(x=5 sin theta)#
Using the above relationship we obtain
#color(blue)(dx = 5 cos theta" "d theta#
Next, we will rewrite #color(blue)(Exp.1# as
#color(red)(int" " (5 sin theta)^2/(125 cos^3 theta) " " 5 cos theta " "d theta#
Reduce further:
#int " " (5 sin theta)^2/[(5 cos theta)(5^2 cos^2 theta)] " "(5 cos theta) d theta#
#int " " (5 sin theta)^2/[cancel(5 cos theta)(5^2 cos^2 theta)] " "cancel(5 cos theta) d theta#
#int " " [(25 sin^2 theta)/[25 cos^2 theta)] " "d theta#
#int " " [(cancel 25 sin^2 theta)/[cancel 25 cos^2 theta)] " "d theta#
#int " " (tan^2 theta) " "d theta#
Since #color(red)(tan^2 theta = sec^2 theta - 1#,
#int " " (sec^2 theta - 1) " "d theta#
#int " " sec^2 theta " "d theta - int" " 1 " "d theta#
#rArr tan theta - theta + C# #color(blue)(Exp.2#
Using x-substitution
#tan theta = x/sqrt(25-x^2)# and #theta = sin^(-1)(x/5)#
Write #color(blue)(Exp.2# as
#color(blue)(x/[sqrt(25-x^2)]-sin^(-1)(x/5)+C#
