xsin^{2n-1}x/cos^{2n+1}x[0,π/4]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\frac{π}{2}} \frac{x \cos^{p-1} x}{\sin^{p+1} x}dx=\frac{π}{2p}\sec \frac{πp}{2}  (|p| \lt 1)\\
&(2) \displaystyle\int_0^{\frac{π}{4}} \frac{x \sin^{p-1} x}{\cos^{p+1} x}dx=\frac{π}{4p}-\frac{1}{2p}β\left(\frac{p+1}{2}\right)  (p \gt -1)\\
&(3) \displaystyle\int_0^{\frac{π}{4}} \frac{x \sin^{2n-1} x}{\cos^{2n+1} x}dx=\frac{π}{8n}\{1-(-1)^n\}-\frac{1}{2n}\displaystyle\sum_{k=0}^{n-1}\frac{(-1)^k}{2n-2k-1}\\
&(4) \displaystyle\int_0^{\frac{π}{4}} \frac{x \sin^{2n} x}{\cos^{2n+2} x}dx=\frac{1}{2(2n+1)}\left\{\frac{π}{2}+(-1)^{n+1}\log 2-\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{n-k}\right\}\\
\end{alignat}ただし、全て \(n \in \mathrm{N}\)











証明の途中、次の積分の結果を用います。(詳細はこちらです。)
\begin{alignat}{2}
&(A) \displaystyle\int_0^{\frac{π}{4}}(\tan x)^μdx=\frac{1}{2}β\left(\frac{μ+1}{2}\right)  (μ \gt -1)\\
&(B) \displaystyle\int_0^{\frac{π}{4}} \tan^{2n} xdx=\frac{π(-1)^n}{4}+\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k-1}  (n \in \mathrm{N})\\
&(C) \displaystyle\int_0^{\frac{π}{4}} \tan^{2n+1} xdx=\frac{(-1)^n \log 2}{2}+\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k}  (n \in \mathrm{N})
\end{alignat}


また、それぞれ、次の微分における等式を用います。
\begin{alignat}{2}
&(1) (\cot^p x)’=p \cot^{p-1}x \cdot \left(-\frac{1}{\sin^2 x}\right)=-p \cdot \frac{\cos^{p-1} x}{\sin^{p+1} x}\\
&(2) (\tan^p x)’=p \tan^{p-1} x \cdot \frac{1}{\cos^2 x}=p \cdot \frac{\sin^{p-1} x}{\cos^{p+1} x}\\
&(3) (\tan^{2n} x)’=2n \cdot \tan^{2n-1} x \cdot \frac{1}{\cos^2 x}=2n \cdot \frac{\sin^{2n-1} x}{\cos^{2n+1} x}\\
&(4) (\tan^{2n+1} x)’=(2n+1) \cdot \tan^{2n} x \cdot \frac{1}{\cos^2 x}=(2n+1) \cdot \frac{\sin^{2n} x}{\cos^{2n+2} x}
\end{alignat}






<証明>

全て、上記の微分における等式を用いて変形。後に部分積分を行います。

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\frac{π}{2}} \frac{x \cos^{p-1} x}{\sin^{p+1} x}dx=-\frac{1}{p}\displaystyle\int_0^{\frac{π}{2}}x(\cot^p x)’dx=-\frac{1}{p}\left\{\left[x(\cot^p x)\right]_0^{\frac{π}{2}}-\displaystyle\int_0^{\frac{π}{2}}\cot^p xdx\right\}\\
&                  =\frac{1}{p}\displaystyle\int_0^{\frac{π}{2}}\cot^p xdx=\frac{1}{p}\displaystyle\int_0^{\frac{π}{2}} (\cos x)^p (\sin x)^{-p}dx\\
&                  =\frac{1}{p} \cdot \frac{1}{2}B\left(\frac{p+1}{2},\frac{1-p}{2}\right)=\frac{1}{2p} \cdot \frac{π}{\cos \frac{πp}{2}}=\frac{π}{2p}\sec \frac{πp}{2}\\
\end{alignat}







\begin{alignat}{2}
&(2) \displaystyle\int_0^{\frac{π}{4}} \frac{x \sin^{p-1} x}{\cos^{p+1} x}dx=\frac{1}{p}\displaystyle\int_0^{\frac{π}{4}}x(\tan^p x)’dx=\frac{1}{p}\left\{\left[x(\tan^p x)\right]_0^{\frac{π}{4}}-\displaystyle\int_0^{\frac{π}{4}}\tan^p xdx\right\}\\
&                  =\frac{1}{p}\left(\frac{π}{4}-\displaystyle\int_0^{\frac{π}{2}}\tan^p xdx\right)=\frac{π}{4p}-\frac{1}{p}\cdot \frac{1}{2}β\left(\frac{p+1}{2}\right)\\
&                  =\frac{π}{4p}-\frac{1}{2p}β\left(\frac{p+1}{2}\right)\\
\end{alignat}







\begin{alignat}{2}
&(3) \displaystyle\int_0^{\frac{π}{4}} \frac{x \sin^{2n-1} x}{\cos^{2n+1} x}dx=\frac{1}{2n}\displaystyle\int_0^{\frac{π}{4}}x(\tan^{2n} x)’dx=\frac{1}{2n}\left\{\left[x(\tan^{2n} x)\right]_0^{\frac{π}{4}}-\displaystyle\int_0^{\frac{π}{4}}\tan^{2n} xdx\right\}\\
&                   =\frac{1}{2n}\left(\frac{π}{4}-\displaystyle\int_0^{\frac{π}{2}}\tan^{2n} xdx\right)=\frac{1}{2n}\left(\frac{π}{4}-\frac{π(-1)^n}{4}-\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k-1}\right)\\
&                   =\frac{π}{8n}\{1-(-1)^n\}-\frac{1}{2n}\displaystyle\sum_{k=0}^{n-1}\frac{(-1)^k}{2n-2k-1}
\end{alignat}







\begin{alignat}{2}
&(4) \displaystyle\int_0^{\frac{π}{4}} \frac{x \sin^{2n} x}{\cos^{2n+2} x}dx=\frac{1}{2n+1}\displaystyle\int_0^{\frac{π}{4}}x(\tan^{2n+1} x)’dx=\frac{1}{2n+1}\left\{\left[x(\tan^{2n+1} x)\right]_0^{\frac{π}{4}}-\displaystyle\int_0^{\frac{π}{4}}\tan^{2n+1} xdx\right\}\\
&                  =\frac{1}{2n+1}\left(\frac{π}{4}-\displaystyle\int_0^{\frac{π}{2}}\tan^{2n+1} xdx\right)=\frac{1}{2n}\left\{\frac{π}{4}-\frac{(-1)^n \log 2}{2}-\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k}\right\}\\
&                  =\frac{1}{2(2n+1)}\left\{\frac{π}{2}+(-1)^{n+1}\log 2-\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{n-k}\right\}
\end{alignat}

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