What is the antiderivative of #(9-x^2)^(1/2)#?

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Answer 1 · 2018-07-10T22:28:08

#\int (9-x^2)^{1/2}\ dx=1/2x\sqrt{9-x^2}+9/2\sin^{-1}(x/3)+C#

Let #x=3\sin\theta\implies dx=3\cos\theta\ d\theta#

#\therefore \int (9-x^2)^{1/2}\ dx#

#= \int (9-(3\sin\theta)^2)^{1/2}\ (3\cos\theta\ d\theta)#

#=3\int (3\cos\theta)\cos\theta\ d\theta#

#=9\int \cos^2\theta\ d\theta#

#=9\int (\frac{1+\cos2\theta}{2})\ d\theta#

#=9/2\int (1+\cos2\theta)\ d\theta#

#=9/2 (\theta+1/2\sin2\theta)+C#

#=9/2 (\sin^{-1}(x/3)+\sin\theta\cos\theta)+C#

#=9/2 (\sin^{-1}(x/3)+\sin\theta\sqrt{1-\sin^2\theta})+C#

#=9/2 (\sin^{-1}(x/3)+x/9\sqrt{9-x^2})+C#

#=1/2x\sqrt{9-x^2}+9/2\sin^{-1}(x/3)+C#