## Calculus, 10th Edition (Anton)

$\int \frac{x^2}{\sqrt {16-x^2}}dx = 8sin^{-1}(\frac{x}{4}) -\frac{x\sqrt {16-x^2}}{2} + c$
Begin by letting $x = 4sin(u)$ $dx = 4cos(u)du$ Substitute: $\int \frac{(4sin(u))^2}{\sqrt {16-(4sin(u))^2}}4cos(u)du$ Simplify: $\int \frac{16sin^2(u)}{\sqrt {16(1-sin^2(u))}}4cos(u)du$ Use the trigonometric identity: $cos^2(u) = 1 - sin^2(u)$: $\int \frac{16sin^2(u)}{\sqrt {16cos^2(u)}}4cos(u)du$ Simplify: $\int 16sin^2(u)du$ Use the identity $sin^2(u) = \frac{1-cos(2u)}{2}$: $\int 8 - 8cos(2u))du$ Integrate: $8u -4sin(2u) + c$ Use the identity $sin(2u) = 2sin(u)cos(u)$: $8u -8sin(u)cos(u) + c$ Substitute back: Since $x = 4sin(u)$, then $u = sin^{-1}(\frac{x}{4})$ $8sin^{-1}(\frac{x}{4}) -8sin(sin^{-1}(\frac{x}{4}))cos(sin^{-1}(\frac{x}{4})) + c$ Simplify: $8sin^{-1}(\frac{x}{4}) -\frac{x\sqrt {16-x^2}}{2} + c$