sin^{2}x/√(1+sin^{2}x)[0,π/2]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\frac{π}{2}}\frac{\sin^2 x}{\sqrt{1+\sin ^2 x}}dx=\sqrt{2}\boldsymbol{E}\left(\frac{\sqrt{2}}{2}\right)-\frac{1}{\sqrt{2}}\boldsymbol{K}\left(\frac{\sqrt{2}}{2}\right)\\
&(2) \displaystyle\int_0^{\frac{π}{2}}\frac{\cos^2 x}{\sqrt{1+\sin ^2 x}}dx=\sqrt{2}\left\{\boldsymbol{K}\left(\frac{\sqrt{2}}{2}\right)-\boldsymbol{E}\left(\frac{\sqrt{2}}{2}\right)\right\}\\
\end{alignat}









<証明>

どちらも \(\displaystyle x=\frac{π}{2}-t\) と置きます。\((dx=dt)\)
\begin{alignat}{2}
&(1) \displaystyle\int_0^{\frac{π}{2}}\frac{\sin^2 x}{\sqrt{1+\sin^2 x}}dx=\displaystyle\int_{\frac{π}{2}}^0 \frac{\cos^2 t}{\sqrt{1+\cos^2 t}}(-dt)=\displaystyle\int_0^{\frac{π}{2}}\frac{\cos^2 t}{\sqrt{1+\cos^2 t}}dt\\
&                    =\displaystyle\int_0^{\frac{π}{2}}\frac{1-\sin^2 t}{\sqrt{2-\sin^2 t}}dt=\displaystyle\int_0^{\frac{π}{2}}\frac{(2-\sin^2 t)-1}{\sqrt{2-\sin^2 t}}dt\\
&                    =\displaystyle\int_0^{\frac{π}{2}}\sqrt{2- \sin^2 t}dt -\displaystyle\int_0^{\frac{π}{2}} \frac{1}{\sqrt{2-\sin^2 t}}dt\\
&                    =\sqrt{2}\displaystyle\int_0^{\frac{π}{2}}\sqrt{1-\frac{1}{2}\sin^2 t}dt-\frac{1}{\sqrt{2}}\displaystyle\int_0^{\frac{π}{2}}\frac{1}{\sqrt{1-\frac{1}{2}\sin^2 t}}dt\\
&                    =\sqrt{2}\boldsymbol{E}\left(\frac{\sqrt{2}}{2}\right)-\frac{1}{\sqrt{2}}\boldsymbol{K}\left(\frac{\sqrt{2}}{2}\right)
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{2}}\frac{\sin^2 x}{\sqrt{1+\sin ^2 x}}dx=\sqrt{2}\boldsymbol{E}\left(\frac{\sqrt{2}}{2}\right)-\frac{1}{\sqrt{2}}\boldsymbol{K}\left(\frac{\sqrt{2}}{2}\right)$$








\begin{alignat}{2}
&(2) \displaystyle\int_0^{\frac{π}{2}}\frac{\cos^2 x}{\sqrt{1+\sin^2 x}}dx=\displaystyle\int_{\frac{π}{2}}^0 \frac{\sin^2 t}{\sqrt{1+\cos^2 t}}(-dt)=\displaystyle\int_0^{\frac{π}{2}}\frac{\sin^2 t}{\sqrt{1+\cos^2 t}}dt\\
&                    =-\displaystyle\int_0^{\frac{π}{2}}\frac{-\sin^2 t}{\sqrt{2-\sin^2 t}}dt=-\displaystyle\int_0^{\frac{π}{2}}\frac{(2-\sin^2 t)-2}{\sqrt{2-\sin^2 t}}dt\\
&                    =-\displaystyle\int_0^{\frac{π}{2}}\sqrt{2- \sin^2 t}dt +2\displaystyle\int_0^{\frac{π}{2}} \frac{1}{\sqrt{2-\sin^2 t}}dt\\
&                    =\sqrt{2}\displaystyle\int_0^{\frac{π}{2}}\sqrt{1-\frac{1}{2}\sin^2 t}dt+{\sqrt{2}}\displaystyle\int_0^{\frac{π}{2}}\frac{1}{\sqrt{1-\frac{1}{2}\sin^2 t}}dt\\
&                    =-\sqrt{2}\boldsymbol{E}\left(\frac{\sqrt{2}}{2}\right)+\sqrt{2}\boldsymbol{K}\left(\frac{\sqrt{2}}{2}\right)=\sqrt{2}\left\{\boldsymbol{K}\left(\frac{\sqrt{2}}{2}\right)-\boldsymbol{E}\left(\frac{\sqrt{2}}{2}\right)\right\}
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{2}}\frac{\cos^2 x}{\sqrt{1+\sin ^2 x}}dx=\sqrt{2}\left\{\boldsymbol{K}\left(\frac{\sqrt{2}}{2}\right)-\boldsymbol{E}\left(\frac{\sqrt{2}}{2}\right)\right\}$$

“sin^{2}x/√(1+sin^{2}x)[0,π/2]などの定積分” への2件の返信

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