How to integrate a binomial function 
Evaluate the following integral.
$$\int_{0}^1\displaystyle{207 \choose 7}{x^{200}(1-x)^7}\, dx$$


My attempt was a lengthy one. I opened the integral using binomial expansion and got $7$ different terms which I integrated but one thing that strikes me was since the integral is from $0$ to $1$ and if I replace $x$ by $1-x$ and add the two integrals I might end up with something simpler.

I am not able to implement it or rather I'm struck with it so any help to make this integral simpler.

Edit l know it can be solved by by parts. My query is

$$\int_{0}^1\displaystyle{207 \choose 7}{x^{200}(1-x)^7}\, dx$$ replacing x by 1-x
$$\int_{0}^1\displaystyle{207 \choose 7}{(1-x)^{200}(x)^7}\, dx$$
On adding

2I=$$\int_{0}^1\displaystyle{207 \choose 7}{x^{200}(1-x)^7}+\,^1\displaystyle{207 \choose 200}{x^{200}(1-x)^7} dx$$,
Now since they have become two equidistant terms of binomial expansion (1-x)^2007 can I directly write the sum
 A: I am not sure the text is correct. Assuming that the integral is
$$\binom{207}{7}\int_0^1x^{200}(1-x)^7dx=\frac{207!}{7!200!}\frac{\Gamma(201)\Gamma(8)}{\Gamma(201+8)}=\frac{207!}{7!200!}\frac{200!7!}{208!}=\frac{1}{208}$$
If the text is correct, $\int_0^7 f(x)dx \approx -2.86\times 10^{185}$...(result by the calculator....of course) good luck!
A: Folllowing @zwim's hint, by parts
$$I:=\int_0^1x^{200}(1-x)^7\,dx=\left.\frac{x^{201}(1-x)^7}{201}\right|_0^1+\frac7{201}\int_0^1x^{201}(1-x)^6\,dx\\=\frac7{201}\int_0^1x^{201}(1-x)^6\,dx.$$
Similarly,
$$I=\frac{7\cdot6}{201\cdot202}\int_0^1x^{202}(1-x)^5\,dx=\cdots\\\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{201\cdot202\cdot203\cdot204\cdot205\cdot206\cdot207}\int_0^1x^{207}\,dx.$$
A: Using the incomplete beta function
$$I_{m,n}=\binom{m+n}{n}\int x^m(1-x)^n\,dx=\binom{m+n}{n}B_x(m+1,n+1)$$
$$J_{m,n}=\binom{m+n}{n}\int_0^1 x^m(1-x)^n\,dx=\binom{m+n}{n}\left(B_1(m+1,n+1)-B_0(m+1,n+1)\right)$$ Going from beta to gamma function
$$J_{m,n}=\binom{m+n}{n}\frac{\Gamma (m+1) \Gamma (n+1)}{\Gamma (m+n+2)}=\frac 1 {m+n+1}$$
A: $$\int_0^1x^{200}(1-x)^7dx=\frac{1}{208}$$

This belong to series $$(x+(1-x))^207=1$$
.there are total 208 terms in the series since there are 104 pairs whose sum is equal to mid two terms . So total 104 terms with equal sum. In which the integral we want I.e 2I exists as one of the pair

1= 104×2I   giving I as 1/208
