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$$\int x\times\sqrt{8-x^2} \,dx = \,?$$ I got to this: $$\int\sqrt{8x^2-x^4} \,dx$$ or: $$\int\frac{8x-x^3}{\sqrt{8+x^2}}\, dx$$

I don't know how to integrate neither. If possible, no $\sin$ \ $\cos$ \ etc.

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Have you seen the method of substitution before, sometimes called $u$-substitution? – Graphth Nov 3 '12 at 20:11

2 Answers 2

up vote 5 down vote accepted

Make the substitution $u=8-x^2$, and you’ll get an easy integration.

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Or directly:

$$\int x\sqrt{8-x^2}\,dx=-\frac{1}{2}\int\sqrt{8-x^2}\,d(8-x^2)=-\frac{1}{2}\frac{2}{3}(8-x^2)^{3/2}+C$$

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