The value of $ \int _{0}^{1}x^{99}(1-x)^{100}dx $ is The value of $\int _{0}^{1}x^{99}(1-x)^{100}dx $ is
Not able to do. I'm trying substituton. But clear failure. Please help.
 A: You can just repeatedly integrate by parts:
\begin{align}
\int_0^1 x^{99}(1 - x)^{100}\,dx &= {99 \over 101} \int_0^1 x^{98}(1 - x)^{101}\,dx \\
&= {99 \over 101} {98 \over 102} \int_0^1 x^{97}(1 - x)^{102}\,dx \\
&= {99 \over 101} {98 \over 102} {97 \over 103} \int_0^1 x^{96}(1 - x)^{103}\,dx \\
&\cdots \\
&= 99! {100! \over 199!} \int (1 - x)^{199}\,dx \\
&= {99! \,100! \over 200!}
\end{align}
A: Denote $B(r,s) = \int_0^1 x^{r-1}(1-x)^{s-1}\ dx$. Then your integral is $B(100,101)$, which is just the Beta Function.
So your integral is just $\frac{99! 100!}{200!}$.
A: $\int_0^1x^{99}(1-x)^{100}dx=B(100,101)=\frac{99!100!}{200!}$
A: Let
$$I(n,m)=\int_0^1 x^n(1-x)^mdx$$
then by integration by parts we get
$$I(n,m)=\frac m{n+1}I(n+1,m-1)$$
and by induction we have
$$I(n,m)=\frac{m!n!}{(m+n)!}I(n+m,0)=\frac{m!n!}{(m+n+1)!}$$
A: HINT:
Use $I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$
$I+I=\int_a^bf(x)\ dx+\int_a^bf(a+b-x)\ dx$
Then set $x=\sin^2\theta$
A: This answer is due to T. Bayes known as Bayes Billiard Balls where he tells nice stories to establish:
$$\int_0^1 {n\choose{k}}x^k(1-x)^{n-k}dx=\frac{1}{n+1}$$
Therefore 
$$\int_0^1 {199\choose{99}}x^{99}(1-x)^{199-99}dx=\frac{1}{199+1}$$thus
$$\int_0^1 x^{99}(1-x)^{100}dx=\frac{99!100!}{199!}\frac{1}{200}=\frac{99!100!}{200!}$$
