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