Substitution Problem in definite Integrals I am given
$$\frac{\int_0^1(1-x^{50})^{100 }dx}{ \int_0^1(1-x^{50})^{101}dx}$$
I substituted $$1-x^{50}=k$$ and
Then $$-50x^{49}dx=dk$$
So since each Integral will be divided by $$ -50x^{49}$$ to substitute for $k$ can it get cancelled from both numerator and denominator ??
Leaving integral to be 
$$\frac{\int_1^0(t)^{100 }dx}{ \int_1^0(t)^{101} dx}$$
Which finally comes to be $102/101$??

Try to do integration by part: for any $n\ge 1$, 
$$\int_0^1 (1-x^{50})^n dx = x (1-x^{50})^n \bigg|_0^1 + \int_0^1 x n(1- x^{50})^{n-1} 50 x^{49} dx$$
$$ = 50n \int_0^1 x^{50} (1-x^{50})^{n-1} dx = -50n \int_0^1 (1-x^{50})^n dx + 50n\int_0^1 (1-x^{50})^{n-1} dx$$
$$\Rightarrow \int_0^1 (1-x^{50})^n dx = \frac{50n}{50n+1}\int_0^1 (1-x^{50})^{n-1}dx.$$
$$\Rightarrow \frac{\int_0^1 (1-x^{50})^{n-1} dx}{\int_0^1 (1-x^{50})^{n} dx} = \frac{50n+1}{50n}. $$
Now put $n = 101$. 
