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How do I integrate the following:
∫3x/√(1-x²)dx
I tried integration by parts where u=3x, du=3, v=(1-x²)^-½, dv=-½(1-x²)^(-3/2)-2x(1-x²)^-½
it works out ok, but I get stuck with the ∫vdu part which is ∫3(1-x²)^-½
Any help?
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Not entirely sure that this works, by try substituting x = sin u in there.
That √(1-x²) is begging for some kind of trigonometric substitution.
Why did the vector cross the road?
It wanted to be normal.
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Set u = 1 - x^2, then you get
It should be straightforward from there.
Wrap it in bacon
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