Value of $5050\cdot\int^1_0(1-x^{50})^{100} dx/\int^1_0(1-x^{50})^{101} dx$ Please help me find the value of the following integral:
$$\frac{(5050)\int^1_0(1-x^{50})^{100} dx}{\int^1_0(1-x^{50})^{101} dx}$$
I tried solving both numerator and denominator via by-parts but it isn't giving me a conclusive solution. Any other suggestions?
 A: Let's try integrating by parts and see what happens.
$$ \int_0^1 (1-x^n)^m \, dx = \left[ x(1-x^n)^m \right]_0^1 - \int_0^1 -nmx^n(1-x^n)^{m-1} \, dx = 0+ nm\int_0^1 x^n(1-x^n)^{m-1} \, dx. $$
We can fiddle with the right-hand side to get it into a more familiar form:
$$ \int_0^1 x^n(1-x^n)^{m-1} \, dx = \int_0^1 \left( 1 -(1-x^n) \right) (1-x^n)^{m-1} \, dx = \int_0^1 (1-x^n)^{m-1} \, dx - \int_0^1 (1-x^n)^{m} \, dx. $$
Collecting the copies of the integrals together, we find
$$ (1+nm)\int_0^1 (1-x^n)^m \, dx = nm \int_0^1 (1-x^n)^{m-1} \, dx, $$
or
$$ \frac{\int_0^1 (1-x^n)^{m-1} \, dx}{\int_0^1 (1-x^n)^m \, dx} = \frac{1+nm}{nm}. $$
Putting $n=50$, $m=101$, the right-hand side becomes $ 5051/5050 $.
A: one way to solve it besides the change of variable,is by using the binomial theorem in both denominator and enumerator.
$\sum_{n=0}^{100}\binom{100}{n}(x^{50})^{100-n}(-1)^n=(1-x^{50})^{100}$  
$\binom{n}{k}=\frac{n!}{k!(n-k)!}$
Use the linearity of the integral and integrate each term of the series.
