How do you integrate #int 1/sqrt(16-x^2)# by trigonometric substitution?

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Answer 1 · 2016-11-21T04:56:44

#=sin^(-1)(1/4x)+C#

#int1/(sqrt(16-x^2))dx#

#int1/(sqrt(16(1-x^2/16)))dx#

#int1/(4sqrt(1-(x/4)^2))dx#

#1/4int1/(sqrt(1-(x/4)^2))dx#

Let #u = 1/4x#
#(du)/(dx)=1/4#
#4du=dx#

#1/4int1/sqrt(1-u^2)4du#

#int1/sqrt(1-u^2)du#

#=sin^(-1)u+C#

Substitute #u# back in:
#=sin^(-1)(1/4x)+C#