Integration Problem

Use $\int_0^1(1-x^{10})^p \cdot dx =$ $1 - \frac {p}{11}+\frac{p(p-1)}{2!}\frac{1}{21}-\frac{p(p-1)(p-2)}{3!}\frac {1}{31}+\frac {p(p-1)(p-2)(p-3)}{4!}\frac {1}{41}-$ . . . and so on as the value obtained at x=1 of the term by term integration of the binomial expansion . This gives p=9 for numerator and p=10 for denominator. Next to find $100 \times \dfrac{\int_0^1(1-x^{10})^9 \cdot dx}{\int_0^1(1-x^{10})^{10}\cdot dx}$ integrate the denominator by parts. $\int_0^1(1-x^{10})^{10}\cdot1\cdot dx = (1-x^{10})^{10}\cdot x|_0^1-\int_0^110(1-x^{10})^9(-10x^9)\cdot xdx$ $=0+100\int_0^1(1-x^{10})^9[1-(1-x^{10})]\cdot dx$ which upon transposing gives the result as 101.