Step 2: Evaluate the second integral: $$ I_2 = \int \frac{3x}{\sqrt{16 - x^2}} dx $$
Let $u = 16 - x^2$. Then $du = -2x \, dx$, so $x \, dx = -\frac{1}{2} du$.
$$ I_2 = \int \frac{3}{\sqrt{u}} \left(-\frac{1}{2}\right) du = -\frac{3}{2} \int u^{-1/2} du $$
$$ I_2
Show more…