$\int x\times\sqrt{8-x^2} \,dx = \,?$ 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.
$\int x\times\sqrt{8-x^2} \,dx = \,?$ 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.
Make the substitution $u=8-x^2$, and you’ll get an easy integration.
Or directly:
$\int x\sqrt{8-x^2}\,dx=-\frac{1}{2}\int\sqrt{8-x^2}\,d(8-x^2)=-\frac{1}{2}\frac{2}{3}(8-x^2)^{3/2}+C$