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$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx =$ $ ?$

I tried putting $x=sin(a)$ but I could do nothing about the 100 and 101 powers, they made the integration not solvable for me.

How should I do or approach this question?

$ dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx$

Let $displaystyle I_n = int_0^1 (1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n+1dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)(1-x^m)^n dx$

$implies displaystyle I_n+1 = int_0^1 (1-x^m)^n dx- int_0^1x^m(1-x^m)^n dx$ ...[*]

$implies displaystyle I_n+1 = I_n- int_0^1x^m(1-x^m)^n dx$

We will integrate $displaystyle int_0^1x^m(1-x^m)^n dx$ via integration by parts. Watch closely, this is a little tricky.

$displaystyle int_0^1x^m(1-x^m)^n dx=int_0^1xcdot x^m-1(1-x^m)^ndx$

Let $u = x implies du = dx$

And $dv = x^m-1(1-x^m)^n dx$

Let $y = (1-x^m) implies dy = -mx^m-1 dx implies x^m-1 dx = -dfracdym$

$displaystyle v = intx^m-1(1-x^m)^n dx = int-dfracy^nmdy = dfrac-y^n+1m(n+1)=dfrac-(1-x^m)^n+1m(n+1)$

$displaystyle int_0^1x^m(1-x^m)^n dx$

$=displaystyle left[-dfracx(1-x^m)^n+1m(n+1)right]_0^1+dfrac1m(n+1)int_0^1(1-x^m)^n+1dx$

$= dfracI_n+1m(n+1)$

Substituting this result into [*]

$I_n+1 = I_n - dfracI_n+1m(n+1)$

$implies left[1+dfrac1m(n+1)right]=dfracI_nI_n+1$

$implies dfracI_nI_n+1 = dfracm(n+1)+1m(n+1)$

Putting $m = 50$ and $n = 100$, we have

$displaystyle dfracint_0^1(1-x^50)^100dxint_0^1(1-x^50)^101dx=dfrac50times101+150times 101=dfrac50515050$

Let $a$, $b>0$. Then, substituting $t=x^a$,
$$I_a,b=
int_0^1(1-x^a)^b,dx=frac1aint_0^1(1-t)^bt^1/a-1,dt=fracB(b+1,1/a)a$$
where $B$ denotes the Beta function. But the beta function is expressible
in terms of the Gamma function so that
$$fracB(b+1,1/a)a=fracGamma(1/a)Gamma(b+1)aGamma(1/a+b+1).$$

Therefore
$$fracI_a,bI_a,b+1=fracGamma(b+1)Gamma(1/a+b+2)Gamma(b+2)
Gamma(1/a+b+1)=frac1/a+b+1b+1.$$
Now let $a=50$ and $b=100$.