In a solutions manual I see they integrate this $\frac{1}{2}r(4-r^2)^2$ and in the next this is $-\frac{1}{12}(4-r^2)^3$. Is this possible without working out the parentheses?
Can someone explain this step? Thanks!
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Let $u=4-r^2$. Or else guess the answer will look like $k(4-r^2)^3$, and differentiate to see what $k$ should be. – André Nicolas Apr 12 '14 at 20:00
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Yes, definitely. We are given
$$\int \dfrac{1}{2}r(4 - r^2)^2\,dr$$
Set $u = 4 - r^2$. Then, $du = -2r\,dr$, which gives
$$-\dfrac{1}{4} \int u^2\,du = -\dfrac{1}{12} u^3 + \mbox{C}$$
Thus, we have $-\frac{1}{12}(4 - r^2)^3 + \mbox{C}$.
NasuSama
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