证明一积分递推关系式

╰☆ヾo.海x

╰☆ヾo.海x

已会

其妙

作差，提取公因式后，约分，还比较容易。

$2(n+1)(I_{n+1}-I_n)=2(n+1)\int_{\frac{\pi}2}^{x}\dfrac{\cos^{2n+1}t(\cos^2t-1) }{\sin t}\mathrm{d}t=2(n+1)\int_{\frac{\pi}2}^{x}\cos^{2n+1}t\mathrm{d}(\cos t)=\cos^{2n+2}t|_{\frac{\pi}2}^{x}=\cos^{2n+2}x$

hbghlyj

\[\mathrm{B}_{x}\left(a,b\right)=\int_{0}^{x}t^{a-1}(1-t)^{b-1}\mathrm{d}t,\]
\[I_{x}\left(a,b\right)=\mathrm{B}_{x}\left(a,b\right)/\mathrm{B}\left(a,b\right),\]
Beta in terms of sine and cosine
\[\int_{\pi/2}^x\cos ^{2 a+1}(t)\sin^{2b+1} (t)\mathrm{d}t=-\frac12B_{\cos^2(x)}(a,b)\]
\[\int_{\pi/2}^x\frac{\cos ^{2 n+1}(t)}{\sin (t)}dt=-\frac{1}{2} B_{\cos^2(x)}(n+1,0)\]
§8.17(iv) Recurrence Relations
DLMF
\[(a+b)I_{x}\left(a,b\right)=aI_{x}\left(a+1,b\right)+bI_{x}\left(a,b+1\right),\]