- #1
neshepard
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Homework Statement
∫√(9-x^2)
Homework Equations
The Attempt at a Solution
(9-x^2)^1/2
1/2(9x-(x^3/3))^3/2
(9x-(x^3/3))/3
But this is wrong and I can't see where or how.
Where or how do you get sin and cos into this equation?
∫(9-3sin(u)^2)^1/2*3cos(u)du
∫(9-9sin^2(u))^2)^1/2*3cos(u)du
3∫(1-sin^2(u))^1/2*cos(u)*du
9∫(cos^2(u))^1/2*cos(u)*du
2(sin^2(sin^-1(x/3)))^3/2
neshepard said:Where or how do you get sin and cos into this equation?
this is a lot more intuitive
9/2[arcsin(x/3)+sin(arcsin(x/3)cos(arcsin(x/3)]
9/2[arcsin(x/3)+(x√(9-x^2))/3]+C
Dickfore said:Inverse functions.
The questions you are asking have nothing to do with Calculus, though. Are you sure you had a pre Calculus course?
Dickfore said:Yes, he made an error with a factor of 1/3. Good thing you caught it.
Ah, that explains everything.neshepard said:community college