# Evaluating $\int y^4(1-y)^3 dy$ using integration by parts

Here is the function which could easily be solved using expansion method but how could I solve it using integration by parts

$$\int y^4(1-y)^3 dy$$

The problem is, when I apply integration by parts to solve it, it is never ending solution and I am not able to get the answer.

For example, I let $u = (1-y)^3$ and $dv = (y^4)$, so $du = 3(1-y)^2$ and $v = \dfrac{y^5}{5}$

When I apply the Integration by Parts formula,

$$uv - \int v du$$

I got the kind of same equation as I started with, so I need to apply integration by parts once again, and then again. How many times is it required to apply before I get the answer ?

• Dear Jason, welcome to the site. Kindly read here (meta.math.stackexchange.com/questions/107/…) on how to typeset math on this website. It is not easy to read what you have written without typeset.
– user17762
May 14, 2012 at 22:59
• @Marvis I suggested an edit - it's pending now.
– Joe
May 14, 2012 at 23:00
• Two should be sufficient, with a little algebra afterwards. May 14, 2012 at 23:03
• Bouncing edits back and forth with @Marvis is fun.
– Joe
May 14, 2012 at 23:06
• By the way, showing work besides your variables is never a bad idea. Perhaps you got to the end of the second by parts and had difficulties simplifying - it would let others know how to help you more directly.
– Joe
May 14, 2012 at 23:09

$$I(4,3) = \int y^4 (1-y)^3 \mathrm{d}y$$
You were heading in the right direction, i.e. $\mathrm{d}v=y^4 \Rightarrow v = \frac{y^5}{5}$
\begin{align*} I(4,3) &= \frac{y^5(1-y)^3}{5} + \frac{3}{5} \int y^5 (1-y)^2 \mathrm{d}y\\ &= \frac{y^5(1-y)^3}{5} + \frac{3}{5} I(5,2)\\ \end{align*}
Similarly use $\mathrm{d}v=y^5$, $u=(1-y)^2$ to evaluate $I(5,2)$
\begin{align*} I(5,2) &= \frac{y^6(1-y)^2}{6} +\frac{1}{3} I(6,1)\\ &= \frac{y^6(1-y)^2}{6} + \frac{1}{3}\left(\frac{y^7}{7} - \frac{y^8}{8}\right)\\ I(4,3) &= \frac{y^5(1-y)^3}{5} + \frac{1}{10}y^6(1-y)^2+\frac{1}{5}\left(\frac{y^7}{7} - \frac{y^8}{8}\right)+ Constant \end{align*}