Let #x = 5sin(u) implies dx = 5cos(u)du#
Integral becomes
#int (25sin^2(u)5cos(u))/(sqrt(25-25sin^2(u)))du#
#75int (sin^2(u)cos(u))/(5sqrt(1-sin^2(u)))du#
Recall that #sin^2x+cos^2x = 1 implies cos^2x = 1-sin^2x#
#25int (sin^2(u)cos(u))/sqrt(cos^2(u))du = 25intsin^2(u)du#
Given #cos(2theta) = 1 - 2sin^2(theta)#
We have that #sin^2(u) = 1/2(1 - cos(2u))#
Integral becomes
#(25)/2 int 1 - cos(2u) du = 25/2[u - 1/2sin(2u)] + C#
#x=5sin(u) implies u = sin^(-1)(x/5)#
So our answer is
#25/2[sin^(-1)(x/5) - 1/2sin(2sin^(-1)(x/5))]+C#
As is, I don't really want to deal with that second term, so what we do is go back a step and expand #sin(2u) = 2sin(u)cos(u)#
#25/2[sin^(-1)(x/5) - sin(sin^(-1)(x/5))cos(sin^(-1)(x/5))]+C#
Obviously sin and arcsin will cancel out but how do we deal with cos and arcsin? My preferred method is using a simple right angled triangle.
Consider #y=sin^(-1)(x)#
#implies x = sin(y)#
Given sine is defined as opposite/hypotenuse this means that in our right angled triangle we have opposite side to the angle #y# of length #x# and hypotenuse of length 1. By Pythagoras the adjacent side will therefore be #sqrt(1-x^2)#.
#cos(y) = cos(sin^(-1)(x)) = ("Adjacent")/("Hypotenuse") = sqrt(1-x^2)#
So #cos(sin^(-1)(x/5)) = sqrt(1-(x/5)^2) = 1/5sqrt(25-x^2)#
Answer is
#25/2[sin^(-1)(x/5) - (x/5)(1/5sqrt(25-x^2))] + C#
#25/2[sin^(-1)(x/5) -x/(25)sqrt(25-x^2)]+C#
#25/2sin^(-1)(x/5) - x/2sqrt(25-x^2)+C#
