Answer to Question #103217 in Calculus for Patrick Kariuki

Question

Integrate x²dx/(9-x²)½

Answer 1

$$\int\frac{x^2}{\sqrt{9-x^2}}dx$$

We will replace the variables:

$x=3\sin{t} -> dx=3\cos{t} dt$.

$\int\frac{27\sin^2{t}\cos{t}}{\sqrt{9-9\sin^2{t}}}dt=\int \frac{27\sin^2{t}\cos{t}}{3\sqrt{1-\sin^2{t}}}dt=\int\frac{9\sin^2{t}\cos{t}}{\cos{t}}dt=9\int\sin^2{t} dt$

Use transformation:

$\sin^2{t}=\frac{1-\cos{2t}}{2}$,

then we get

$9\int \sin^2{t}dt=9\int \frac{1-\cos{2t}}{2}dt=\frac{9}{2}\int (1-\cos{2t})dt=\frac{9}{2}(t-\frac{1}{2}\sin{2t})+C=\frac{9}{2}(t-\sin{t}\cos{t})+C$

where $C$ - contact.

$t=\arcsin{\frac{x}{3}}$

$\sin{t}=\frac{x}{3}$

$\cos{t}=\sqrt{1-\sin^2{t}}=\sqrt{1-\frac{x^2}{9}}$

$\frac{9}{2}(t-\sin{t}\cos{t})+C=\frac{9}{2}(\arcsin{\frac{x}{3}}-\frac{x}{3}\sqrt{1-\frac{x^2}{9}})+C=\frac{9\arcsin{\frac{x}{3}}-x\sqrt{9-x^2}}{2}+C$

Answer:

$\int\frac{x^2}{\sqrt{9-x^2}}dx=\frac{9\arcsin{\frac{x}{3}}-x\sqrt{9-x^2}}{2}+C$