How do I integrate $\int x\sqrt{A^2 - x^2} \mathrm dx$ $$ \int x\sqrt{A^2 - x^2} \mathrm dx $$
I never dealt with 2 variables on integration, what should I do here?
 A: $A$ is not a variable here (you can tell the variable of integration is $x$, by the cue $dx$.)
$$ \int x\sqrt{A^2 - x^2} \mathrm dx = \int x(A^2 - x^2)^{1/2} \,dx$$
Let $\,u = A^2 - x^2.\;$ Then $\,du = -2x dx \implies -\dfrac 12 du = x\,dx$.
Now substitute: $$\int -\dfrac 12 u^{1/2} \,du = -\frac 12 \cdot \frac 23(u)^{3/2} + C$$
Now back substitute and simplify: $$-\frac 13 (A^2 - x^2)^{3/2} + C = -\frac 13(A^2 - x^2) \sqrt{A^2 -x^2} + C$$
A: $$\int x(a^2-x^2)^{1/2}dx=-\frac{1}{2}\int(-2xdx)\sqrt{a^2-x^2}=-\frac{1}{2}\frac{2}{3}(a^2-x^2)^{3/2}+C=$$
$$-\frac{1}{3}(a^2-x^2)\sqrt{a^2-x^2}+C$$
A: Alternative way is using the trigonometric substitution $x=A\sin{t}.$
Then $$dx=A\cos{t}\;dt,\\
{A^2 - x^2}=A^2-A^2{\sin^2{t}}=A^2{\cos^2{t}},$$ therefore
$$\int x\sqrt{A^2 - x^2}\;  dx = \int {A\sin{t}\,A{\cos{t}}}A\cos{t}\;dt =\\
= -A^3 \int {\cos^2{t}\;d(\cos{t})}=-\frac{A^3}{3}\cos^3{t}+C=  -\frac{A^3}{3} \left(1-\sin^2{t} \right)^{\tfrac{3}{2}}+C =\\
=-\frac{A^3}{3}\left({1-\frac{x^2}{A^2}}\right)^{\tfrac{3}{2}}+C=-\frac{1}{3}\left({A^2-{x^2}}\right)^{\tfrac{3}{2}}+C.$$
