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 $102/101$??
 A: 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 $n = 101$. 
A: Recall that the Beta function is $$B\left(p,q\right)=\int_{0}^{1}u^{p-1}\left(1-u\right)^{q-1}du.\,\,\,\,\,\,\,\,\,(1)$$
 Now put $x^{50}=u\rightarrow dx=\frac{1}{50}u^{\frac{1}{50}-1}du
 .$ So$$\int_{0}^{1}\left(1-x^{50}\right)^{100}dx=\frac{1}{50}\int_{0}^{1}u^{1/50-1}\left(1-u\right)^{101-1}du=\frac{1}{50}B\left(\frac{1}{50},101\right).$$
 The other integral is similar. If we wan elaborate a bit more, we can use the identity $$B\left(p,q+1\right)=\frac{q}{p+1}B\left(p,q\right)$$
 and so $$\frac{\int_{0}^{1}\left(1-x^{50}\right)^{100}dx}{\int_{0}^{1}\left(1-x^{50}\right)^{101}dx}=\frac{B\left(\frac{1}{50},101\right)}{B\left(\frac{1}{50},102\right)}=\frac{\frac{1}{50}+101}{101}=\frac{5051}{5050}.$$
 Note that you can't cancels as you said, because in general $$\frac{\int f\left(x\right)dx}{\int g\left(x\right)dx}\neq\int\frac{f\left(x\right)}{g\left(x\right)}dx.
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