Integrate using substitution(not trigonometric) the following. I will share how I did it, even though the answer is not right.
$$\int_{-1}^1 x \sqrt{1-x^2} dx $$ $$ Substituting \space x^2=y $$ $$2xdx=dy \space $$ $$ \space limit:\space 0 \space to \space 1$$ $$ Therefore \space the \space integral \space changes \space to$$ $$ \frac12\int_0^1 \sqrt{1-y} du$$ $$= -\frac13 [(1-u)^{\frac32}]_0^1$$ $$=\frac13$$
After integration if $$ y=x^2$$ is substituted, correct answer 0 is obtained. But shouldn't this give same answer even if it's not re-substituted, given the limits are changed? Can someone correct the mistake?
