# How to solve this integral $\displaystyle \dfrac{\int_{0}^{1}{(1-x^{50})^{100}}dx}{\int_{0}^{1}{(1-x^{50})^{101}}dx} =$? [duplicate]

$$\displaystyle \dfrac{\int_{0}^{1}{(1-x^{50})^{100}}dx}{\int_{0}^{1}{(1-x^{50})^{101}}dx} =$$ $$?$$

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?

## marked as duplicate by StubbornAtom, Lord Shark the Unknown, Namaste integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 12 '18 at 15:39

• Could you put some form of the integral in your question title for potential question disambiguation, please? – Cœur Oct 12 '18 at 6:33

$$\dfrac{\int_{0}^{1}{(1-x^{50})^{100}}dx}{\int_{0}^{1}{(1-x^{50})^{101}}dx}$$

Let $$\displaystyle I_n = \int_{0}^{1} {(1-x^m)^n} dx$$

$$\implies \displaystyle I_{n+1} = \int_{0}^{1} {(1-x^m)^{n+1}}dx$$

$$\implies \displaystyle I_{n+1} = \int_{0}^{1} {(1-x^m)(1-x^m)^n} dx$$

$$\implies \displaystyle I_{n+1} = \int_{0}^{1} {(1-x^m)^n} dx- \int_{0}^{1}{x^m(1-x^m)^n} dx$$ ...[*]

$$\implies \displaystyle I_{n+1} = I_n- \int_{0}^{1}{x^m(1-x^m)^n} dx$$

We will integrate $$\displaystyle \int_{0}^{1}{x^m(1-x^m)^n} dx$$ via integration by parts. Watch closely, this is a little tricky.

$$\displaystyle \int_{0}^{1}{x^m(1-x^m)^n} dx=\int_{0}^{1}{x\cdot x^{m-1}(1-x^m)^n}dx$$

Let $$u = x \implies du = dx$$

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

Let $$y = (1-x^m) \implies dy = -mx^{m-1} dx \implies x^{m-1} dx = -\dfrac{dy}{m}$$

$$\displaystyle v = \int{x^{m-1}(1-x^m)^n} dx = \int{-\dfrac{y^n}{m}}dy = \dfrac{-y^{n+1}}{m(n+1)}=\dfrac{-(1-x^m)^{n+1}}{m(n+1)}$$

$$\displaystyle \int_{0}^{1}{x^m(1-x^m)^n} dx$$

$$=\displaystyle \left[-\dfrac{x(1-x^m)^{n+1}}{m(n+1)}\right]_{0}^{1}+\dfrac{1}{m(n+1)}\int_{0}^{1}{(1-x^m)^{n+1}}dx$$

$$= \dfrac{I_{n+1}}{m(n+1)}$$

Substituting this result into [*]

$$I_{n+1} = I_n - \dfrac{I_{n+1}}{m(n+1)}$$

$$\implies \left[1+\dfrac{1}{m(n+1)}\right]=\dfrac{I_n}{I_{n+1}}$$

$$\implies \dfrac{I_n}{I_{n+1}} = \dfrac{m(n+1)+1}{m(n+1)}$$

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

$$\displaystyle \dfrac{\int_{0}^{1}{(1-x^{50})^{100}}dx}{\int_{0}^{1}{(1-x^{50})^{101}}dx}=\dfrac{50\times101+1}{50\times 101}=\dfrac{5051}{5050}$$

Let $$a$$, $$b>0$$. Then, substituting $$t=x^a$$, $$I_{a,b}= \int_0^1(1-x^a)^b\,dx=\frac1a\int_0^1(1-t)^bt^{1/a-1}\,dt=\frac{B(b+1,1/a)}{a}$$ where $$B$$ denotes the Beta function. But the beta function is expressible in terms of the Gamma function so that $$\frac{B(b+1,1/a)}{a}=\frac{\Gamma(1/a)\Gamma(b+1)}{a\Gamma(1/a+b+1)}.$$

Therefore $$\frac{I_{a,b}}{I_{a,b+1}}=\frac{\Gamma(b+1)\Gamma(1/a+b+2)}{\Gamma(b+2) \Gamma(1/a+b+1)}=\frac{1/a+b+1}{b+1}.$$ Now let $$a=50$$ and $$b=100$$.