#int(2/(sqrt(16-4x^2)# dx=?

Question

1147 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-14T01:55:42

#arcsin(x/2)+C#

derivative of #arcsin(f(x))=(f'(x))/sqrt(1-f(x)^2)#

#int2/sqrt(16-4x^2)dx#
#=int2/(4sqrt(1-x^2/4))dx#
#=int(1/2)/sqrt(1-x^2/4)dx#

using the format of #arcsin(f(x))# derivative, if you let #f(x)=x/2# then #f'(x)=1/2#

substituting:
#=arcsin(x/2)+C#