You have used the formula ∫▒〖1/√(a^2-x^2 ) dx = sin^(-1) (x/a) + C where C is constant. How have you got 9 -3pi/6 pl explain
Ok this is all my work, please let me know what went wrong Im very perplexed at the moment (my TA said this answer was good but unfortunately web homework site does not like it):
3/sqrt(1-x^2)
= 3 ∫1/sqrt(1-x^2) dx
=3sin^-1x+c
f(1/2)=3sin^-1(1/2)
=3(pi/6)+c=9
c=9-3pi/6
final answer f(x)=3sin^-1*x+9-(3pi/6)