Answer
$\int \frac{x^2}{\sqrt {16-x^2}}dx = 8sin^{-1}(\frac{x}{4}) -\frac{x\sqrt {16-x^2}}{2} + c$
Work Step by Step
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:
$8sin^{-1}(\frac{x}{4}) -\frac{x\sqrt {16-x^2}}{2} + c$
