Let's say that #x = 2sin(t)#, then #dx = 2cos(t) dt#
So we have
#int(sin^2(t)/cos^2(t)*dt/(2cos(t))) = 1/2int(tan^2(t)*sec(t)dt)#
However we know that #tan^2(x) = sec^2(x) - 1#, so we can say it's
#int(sec^3(t) - sec(t))dt#
Breaking the integral, we have
#intsec^3(t)dt - intsec(t)dt#
Which are both standard integrals, so we have the integral equal
#(sec(t)tan(t))/2 + ln|sec(t) + tan(t)|/2 - ln|sec(t) + tan(t)| + c#
Or, simplifying
#(sec(t)tan(t) -ln|sec(t)+tan(t)|)/2 + c#
#int(sin^2(t)/cos^2(t)*dt/(2cos(t))) = 1/2int(tan^2(t)*sec(t)dt) = (sec(t)tan(t) -ln|sec(t)+tan(t)|)/4 + c#
But the original integral was in terms of #x#, and since
#x = 2sin(t)#, we have that #t = arcsin(x/2)# so substituting that we have
#int ( x^2 / sqrt(4 - x^2) ) dx = (sec(t)tan(t) -ln|sec(t)+tan(t)|)/4 + c#
#I = int ( x^2 / sqrt(4 - x^2) ) dx = (sec(arcsin(x/2))tan(arcsin(x/2)) -ln|sec(arcsin(x/2))+tan(arcsin(x/2))|)/4 + c#
Or, using some algebra to simplify,
We know that #sin(arcsin(x/2)) = x/2# and that #cos(arcsin(x/2)) = sqrt(1 - x^2/4)#, so we have
#sec(arcsin(x/2)) = 1/cos(arcsin(x/2)) = 1/sqrt(1 - x^2/4) = 2/sqrt(4 - x^2)#
#tan(arcsin(x/2)) = x/(2sqrt(1 - x^2/4)) = x/sqrt(4 - x^2)#
#I = (2x)/(4 - x^2) + 1/4*ln|(2 + x)/sqrt(4 - x^2)| + c#
