Given:
#int1/sqrt(7 - 3x^2)dx#
Divide by #sqrt(3)# so that the integrand is in the form #1/sqrt(a^2 - x^2)#
#1/sqrt(3)int1/sqrt(7/3 - x^2)dx#
Please notice that #a^2 = 7/3#, therefore, #a = sqrt(7/3)#
The reference for Trignometric substitutions says to substitute #x = asin(t)#
let #x = sqrt(7/3)sin(t)#, then #dx = sqrt(7/3)cos(t)dt#
#1/sqrt(3)int(sqrt(7/3)cos(t))/sqrt(7/3 - 7/3sin^2(t))dt#
From the same reference, it says to substitute #a^2cos^2(t)# for #a^2 - a^2sin^2(t)#:
#1/sqrt(3)int(sqrt(7/3)cos(t))/sqrt(7/3cos^2(t))dt = #
#1/sqrt(3)int(sqrt(7/3)cos(t))/(sqrt(7/3)cos(t))dt = #
#1/sqrt(3)intdt = (sqrt(3)(t))/3 + C#
Solve the original substitution for t:
#x = sqrt(7/3)sin(t)#
#sin(t) = sqrt(3/7)x#
#t = sin^-1(sqrt(3/7)x)#
Substitute #sin^-1(sqrt(3/7)x)# for #t#
#int1/sqrt(7 - 3x^2)dx = (sqrt(3)(sin^-1(sqrt(3/7)x)))/3 + C#
