Asked by Hailey
Evaluate the indefinite integral.
∫(arcsinx)^6/(1−x^2)^(1/2)dx
∫(arcsinx)^6/(1−x^2)^(1/2)dx
Answers
Answered by
Steve
let
u = arcsin(x)
du = 1/√(1-x^2) dx
and the integrand is
u^6 du
u = arcsin(x)
du = 1/√(1-x^2) dx
and the integrand is
u^6 du
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