How to handle such integrals

Make the substitution $t=1-x^2$ on top and bottom, and simplify, to get $\frac{\int\limits_0^1 (1-t^{100})^{201} \, dt}{\int\limits_0^1 (1-t^{100})^{202} \, dt}.$ Let $I_{a,b} = \int\limits_0^1 (1-t^a)^b \, dt.$ Let's evaluate $I_{a,b}$ by integrating by parts, letting $u = (1-t^a)^b$ and $dv = dt.$ $I_{a,b} = \int\limits_0^1 (1-t^a)^b \, dt = t(1-t^a)^b \bigg|_0^1 + ab\int\limits_0^1 t^a(1-t^a)^{b-1} \, dt.$ The first term on the right side is zero. Note that if we add $abI_{a,b}$ to the second term, we get $ab \int\limits_0^1 (1-t^a)(1-t^a)^{b-1} \, dt + ab \int\limits_0^1 t^a(1-t^a)^{b-1} \, dt = ab \int\limits_0^1 (1-t^a)^{b-1} \, dt = abI_{a,b-1}.$ That is, $abI_{a,b} + I_{a,b} = ab I_{a,b-1}.$ So $\frac{I_{a,b-1}}{I_{a,b}} = \frac{ab+1}{ab}.$ When $a,b$ are integers (here $a=100,b=202$ ), the numerator and denominator are relatively prime, so the difference is $\fbox{1}.$

The integral $\begin{aligned} I_{m,n} & = \; \int_0^1 \big(1 - (1 - x^2)^m\big)^n\,x\,dx \; = \; \tfrac12\int_0^1 \big(1 - (1 - x)^m\big)^n\,dx \; = \; \tfrac12\int_0^1 (1 - x^m)^n\,dx \\ & = \; \frac{1}{2m}\int_0^1 y^{\frac{1}{m}-1}(1-y)^n\,dy \; = \; \tfrac{1}{2m}B(\tfrac{1}{m},n+1) \end{aligned}$ for any positive integers $m,n$ , using the substitution $y = x^m$ . Thus $\frac{I_{m,n}}{I_{m,n+1}} \; = \; \frac{B(\tfrac{1}{m},n+1)}{B(\tfrac{1}{m},n+2)} \; = \; \frac{\Gamma(\tfrac{1}{m})\Gamma(n+1)\Gamma(\tfrac{1}{m}+n+2)}{\Gamma(\tfrac{1}{m}+n+1) \Gamma(\tfrac{1}{m})\Gamma(n+2)} \; =\; \frac{\tfrac{1}{m}+n+1}{n+1} \; = \; \frac{m(n+1) + 1}{m(n+1)}$ so that $p=m(n+1)+1$ , $q = m(n+1)$ are coprime and $p-q = \boxed{1}$ .