If I(m,n)= _1^0x^m-1(1-x)^n-1dx,m,n0, then I(9,14)+I(10,13) is

Question

If $I(m,n)={\int }_{1}^{0}{x}^{m-1}(1-x{)}^{n-1}dx,m,n>0$, then $I(9,14)+I(10,13)$ is

Accepted Answer

$\begin{gathered} I(m,m)={\int }_{1}^{0}{x}^{m-1}(1-x{)}^{n-1}dx \ x={\sin }^{2}	heta \, dx=2\sin 	heta \cos 	heta d	heta \ I(m,n)=2{\int }_{\pi /2}^{0}(\sin 	heta {)}^{2m-1}(\cos 	heta {)}^{2n-1}d	heta \ I(9,14)+I(10,13)=2{\int }_{\pi /2}^{0}(\sin 	heta {)}^{17}(\cos 	heta {)}^{27}d	heta \ +2{\int }_{\pi /2}^{0}(\sin 	heta {)}^{19}(\cos 	heta {)}^{25}d	heta \ =2{\int }_{\pi /2}^{0}(\sin 	heta {)}^{17}(\cos 	heta {)}^{25}d	heta \ =I(9,13) \end{gathered}$

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