Integration

This question is quite interesting, and yet it is not really hard.
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Observe that the function rooted is a^6 - x^2,
remember any trigconometric function such that g(x)^2 = 1 - f(x)^2 ?
you should target at changing it to be of the form 1 - f(x)^2
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do you know how to solve now^^?
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the key is to put x = a^3 * sin (b),
try it!
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dx = a^3 * cos (b) db
I_6 = INT 1 / sqrt( a^6*cos(b)^2 ) * ( a^3 * cos (b) db ) = b + C
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it is similar for I_n
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put x = a^(n/2) * sin(b), and every thing will be canceled out.
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Remarks: If it is definite integral, say, INT from h to k
u got to change the boundary when you change variable
x from h to k, b from arcsin(h/a^3) to arcsin(k/a^3)

2008-06-23 15:36:50 補充：
!!! y=1/(4a^n -x^2)^(1/2)
oops, sorry that i missed out the 4 before a^n =P

∫1/(y^6 - x^2)^(1/2) dx?
y應該不等於f(x)
你想解 f(x.y)
∫∫1/(y^6 - x^2)^(1/2) dx dy ?