tangur
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\int \frac {x^2}{\sqrt{4x-x^2}}dx
I just want to be sure I'm right on this, complete the square first of all so you get -\int \frac {x^2}{\sqrt{(x-2)^2-4}}dx let u=x-2 thus -\int \frac {(x+2)^2}{\sqrt{u^2-4}}dxthen letu=2sec(\theta)
hence integral becomes -8\int sec^3(\theta)+2sec^2(\theta)+sec(\theta)d\theta
and then solve.
Thanks
I just want to be sure I'm right on this, complete the square first of all so you get -\int \frac {x^2}{\sqrt{(x-2)^2-4}}dx let u=x-2 thus -\int \frac {(x+2)^2}{\sqrt{u^2-4}}dxthen letu=2sec(\theta)
hence integral becomes -8\int sec^3(\theta)+2sec^2(\theta)+sec(\theta)d\theta
and then solve.
Thanks
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