Inverse trig function integrals help please
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Inverse trig function integrals help please

[From: ] [author: ] [Date: 13-02-16] [Hit: ]
Please help thank you!= arcsin(x/3) + C-Study Indefinite integration by substitution from any basic standard textbook in integral calculus.Then you can substitute x by say 3t. dx becomes 3dt. Your expression becomes 1/(3*(1-t^2)^1/2). You will get arcsin(t)+c as the answer.......
∫ 1/(9-x^2)^(1/2)

I know it looks similar to the derivative of arc sin, but I don't know how to solve it. The answer is
arcsin(x/3)+c. Please help thank you!

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∫dx/√(9 - x^2) = ∫dx/3*√(1 - (x/3)^2)

z = x/3 => dz = dx/3

∫dz/√(1 - z^2) = arcsin(z) + C

= arcsin(x/3) + C

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Study Indefinite integration by substitution from any basic standard textbook in integral calculus.
Then you can substitute x by say 3t. dx becomes 3dt. Your expression becomes 1/(3*(1-t^2)^1/2). You will get arcsin(t)+c as the answer.
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