積分ガチャの超級から学ぶ積分のテクニック【全積分ガチャ超級の問題, 解答付き】

$t=\tan\dfrac{x}{2}$とおくと, $\d x=\dfrac{2}{1+t^2}\,\d t$
$$ I_1=\int \frac{\cbk{\frac{2t}{1+t^2}}^2}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}+1}\cdot\dfrac{2}{1+t^2}\,\d t =\int \frac{2t^2}{(1+t^2)^2(1+t)}\,\d t =\int \tbk{\frac{1}{2(1+t)}+\frac{1-t}{2(1+t^2)}+\frac{t-1}{(1+t^2)^2}}\,\d t $$

$\dfrac{1}{(1+t^2)^2}$の積分を変形する.
$$ \int \frac{\d t}{1+t^2}=\frac{t}{1+t^2}+\int \frac{2t^2}{(1+t^2)^2}\,\d t =\frac{t}{1+t^2}+2\int \tbk{\frac{1}{1+t^2}-\frac{1}{(1+t^2)^2}}\,\d t \qquad\therefore \int \frac{\d t}{(1+t^2)^2}=\frac{t}{2(1+t^2)}+\frac{1}{2}\int \frac{\d t}{1+t^2} $$
$$ I_1=-\frac{t}{2(1+t^2)}+\int \tbk{\frac{1}{2(1+t)}-\frac{t}{2(1+t^2)}+\frac{t}{(1+t^2)^2}}\,\d t =-\frac{t}{2(1+t^2)}+\frac{1}{2}\int \frac{\d t}{1+t}-\frac{1}{4}\int \frac{(1+t^2)'}{1+t^2}\,\d t+\frac{1}{2}\int \frac{(1+t^2)'}{(1+t^2)^2}\,\d t $$
$$ I_1=-\frac{1+t}{2(1+t^2)}+\frac{1}{2}\log(1+t)-\frac{1}{4}\log(1+t^2)+C $$
$$ \therefore I_1=-\frac{1}{2}\cbk{1+\tan\frac{x}{2}}\cos^2\frac{x}{2}+\frac{1}{2}\log\left|\cos\frac{x}{2}+\sin\frac{x}{2}\right|+C $$

$x=\cos t$とおくと, $\d x=-\sin t\,\d t$
$$ I_2=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\d t}{(\cos t+1)^3} =\frac{1}{8} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\d t}{\cos^6\frac{t}{2}} =\frac{1}{4} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\d t}{\cos^6 t} $$

$s=\tan t$とおくと, $\d s=(s^2+1)\,\d t$
$$ I_2=\frac{1}{4} \int_{\frac{1}{\sqrt{3}}}^1 (1+s^2)^2\,\d s =\frac{1}{4} \bbk{s+\frac{2}{3}s^3+\frac{1}{5}s^5}_{\frac{1}{\sqrt{3}}}^1 =\frac{7}{15}-\frac{14\sqrt{3}}{135} $$

$y=\sqrt{\dfrac{x-1}{2-x}}$とおくと, $2y\,\d y=\dfrac{\d x}{(2-x)^2}$
$$ I_3=\int_1^\sqrt{3} y\cdot 2y(2-x)^2\,\d y $$

$y^2=\dfrac{1}{2-x}-1\Leftrightarrow 2-x=\dfrac{1}{y^2+1}$より
$$ I_3=\int_1^\sqrt{3} \frac{2y^2}{(y^2+1)^2}\,\d y =\bbk{y\cbk{-\frac{1}{y^2+1}}}_1^\sqrt{3}+\int_1^\sqrt{3} \frac{\d y}{y^2+1} =\bbk{-\frac{y}{y^2+1}+\arctan y}_1^\sqrt{3} $$
$$ \therefore I_3=\frac{2-\sqrt{3}}{4}+\frac{\pi}{12} $$

$$ I_4=\int_0^1 \cbk{\sqrt{x^2+1}-\frac{1}{2}\cdot\frac{2x}{\sqrt{x^2+1}}}\,\d x =\bbk{-\sqrt{x^2+1}}_0^1+\int_0^1 \sqrt{x^2+1}\,\d x $$

$x=\sinh y$とおくと, $\d x=\cosh y\,\d y,\,y=\log(x+\sqrt{x^2-1})$
$$ I_4=1-\sqrt{2}+\int_0^{\log(1+\sqrt{2})} \cosh^2 y\,\d y =1-\sqrt{2}+\int_0^{\log(1+\sqrt{2})} \frac{1+\cosh 2y}{2}\,\d y =1-\sqrt{2}+\bbk{\frac{1}{2}y+\frac{1}{4}\sinh 2y}_0^{\log(1+\sqrt{2})} $$
$$ I_4=1-\sqrt{2}+\frac{1}{2}\log(1+\sqrt{2})+\frac{1}{4}\cdot\frac{(1+\sqrt{2})^2-(1+\sqrt{2})^{-2}}{2} =1-\frac{\sqrt{2}}{2}+\frac{1}{2}\log(1+\sqrt{2}) $$

King Propertyで$y=\pi-x$とおいて, 置換前と後の2個の積分を足す.
$$ I_5=\int_0^\pi \frac{(\pi+2y) (\pi-y)^2\sin y}{\sqrt{3-2\sin^2 y}}\,\d y =\frac{1}{2}\int_0^\pi \frac{\{(3\pi-2y) y^2+(\pi+2y) (\pi-y)^2\}\sin y}{\sqrt{3-2\sin^2 y}}\,\d y =\frac{\pi^3}{2}\int_0^\pi \frac{\sin y}{\sqrt{3-2\sin^2 y}}\,\d y $$

$t=\cos y$とおくと, $\d t=-\sin y\,\d y$
$$ I_5=\frac{\pi^3}{2}\int_{-1}^1 \frac{\d t}{\sqrt{1+2t^2}} =\pi^3 \int_0^1 \frac{\d t}{\sqrt{1+2t^2}} =\frac{\pi^3}{\sqrt{2}} \int_0^1 \frac{(\sqrt{2} t)'}{\sqrt{1+(\sqrt{2} t)^2}}\,\d t =\frac{\pi^3}{\sqrt{2}} \bbk{\arsinh{\sqrt{2} t}}_0^1 $$
$$ \therefore I_5=\frac{\pi^3}{\sqrt{2}}\log(\sqrt{2}+\sqrt{3}) $$

$$ I_6=\int_0^{2\pi} \cbk{\sum_{k=1}^n k\sin kx}\cbk{\sum_{l=1}^n l\sin lx}\,\d x =\int_0^{2\pi} \sum_{k=1}^n \sum_{l=1}^n kl \sin kx\sin lx\,\d x =\sum_{k=1}^n \sum_{l=1}^n kl \int_0^{2\pi} \sin kx\sin lx\,\d x $$

$k=l$のとき, $\dps \int_0^{2\pi} \sin kx\sin lx\,\d x=\int_0^{2\pi} \sin^2 kx\,\d x=\bbk{\frac{1-\cos 2kx}{2}}_0^{2\pi}=\pi$
$k\neq l$のとき, $\dps \int_0^{2\pi} \sin kx\sin lx\,\d x=\frac{1}{2}\int_0^{2\pi} \tbk{\cos(k-l)x-\cos(k+l)x}\,\d x=\frac{1}{2}\bbk{\frac{\sin(k-l)x}{k-l}-\frac{\sin(k+l)x}{k+l}}_0^{2\pi}=0$

$$ \therefore I_6=\sum_{k=1}^n \pi k^2=\frac{n(n+1)(2n+1)}{6}\pi $$

$y=\sqrt[3]{x+\sqrt{x^2+1}}$とおくと, $3y^2\,\d y=\dfrac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}\,\d x=\dfrac{y}{\sqrt{x^2+1}}\,\d x$
$y^3=x+\sqrt{x^2+1}$より$y^6=2xy^3+1\Leftrightarrow x=\dfrac{y^6-1}{2y^3}\qquad\therefore\sqrt{x^2+1}=\dfrac{y^6+1}{2y^3}$
$$ I_7=\int \cbk{\frac{y^6+1}{2y^3}}y\cdot\frac{3(y^6+1)}{2y^4}\,\d y =\frac{3}{4}\int \cbk{y^6+2+\frac{1}{y^6}}\,\d y =\frac{3}{4}\cbk{\frac{y^7}{7}+2y-\frac{1}{5y^5}}+C $$
$$ I_7=\frac{3}{4}\cdot\frac{5y^{12}+70y^6-7}{35y^5}+C $$

$y^6=2xy^3+1$であるから$5y^{12}+70y^6-7=5(2xy^3+1)^2+70y^6-7=(20x^2+70)y^6+20xy^3-2=(20x^2+70)(2xy^3+1)+20xy^3-2$
$5y^{12}+70y^6-7=40x(x^2+4)y^3+20x^2+68$となる.
$$ \therefore I_7=\frac{30x(x^2+4)(x+\sqrt{x^2+1})+15x^2+51}{35(x+\sqrt{x^2+1})^{\frac{5}{3}}}+C $$

$2y=x+\dfrac{\pi}{2}$とおく.
$$ I=\int_0^{\frac{\pi}{2}} (1-\cos 2y)^n\,\d y =2^{n+1} \int_0^{\frac{\pi}{2}} \sin^{2n}y\,\d y $$

ウォリス積分より
$$ I=2^{n+1}\cdot\frac{(2n-1)!!}{(2n)!!}\cdot\frac{\pi}{2} =\frac{2^n(2n-1)!!}{(2n)!!}\pi =\dfrac{(2n)!}{2^{n} n!^2}\pi $$

King Propertyより, $y=\pi-x$とおく.
$$ I_9=\int_0^\pi \cos\cbk{\frac{y}{2}}\sqrt{\sin y}\,\d y =\frac{1}{2} \int_0^\pi \cbk{\cos\frac{x}{2}+\sin\frac{x}{2}}\sqrt{2\sin\frac{x}{2}\cos\frac{x}{2}}\,\d x $$

$y=\sin\dfrac{x}{2}-\cos\dfrac{x}{2}$とおくと, $2\,\d y=\cbk{\sin\dfrac{x}{2}+\cos\dfrac{x}{2}}\,\d x,\,2\sin\dfrac{x}{2}\cos\dfrac{x}{2}=1-y^2$
$$ I_9=\int_{-1}^1 \sqrt{1-y^2}\,\d y =\frac{\pi}{2} $$

$t=\sin^4 x$とおくと, $\d t=4\sin^3 x\cos x\,\d x$
$$ I_{10}=\int \frac{e^t (2-t)}{4t^3}\,\d t =\frac{1}{4}\int (2e^t t^{-3}-e^t t^{-2})\,\d t =\frac{1}{4}\int (-e^t t^{-2})'\,\d t =-\frac{e^t}{4t^2}+C $$
$$ I_{10}=-\frac{e^{\sin^4 x}}{4\sin^8 x}+C $$

$$ I_{11}=\int \frac{e^{x}-e^{-x}}{\sqrt{e^{2x}+4e^x+10+4e^{-x}+e^{-2x}}}\,\d x $$

$t=e^x+e^{-x}$とおくと, $\d t=(e^x-e^{-x})\,\d x,\,e^{2x}+e^{-2x}=t^2-2$
$$ I_{11}=\int \frac{\d t}{\sqrt{t^2+4t+8}} =\int \frac{\d t}{\sqrt{(t+2)^2+4}} =\int \frac{(t/2+1)'}{\sqrt{(t/2+1)^2+1}}\,\d t =\arsinh\cbk{\frac{t}{2}+1}+C $$
$$ I_{11}=\arsinh(\cosh x+1)+C=\log\cbk{1+\frac{e^x+e^{-x}}{2}+\sqrt{1+\cbk{\frac{e^x+e^{-x}}{2}}^2}}+C $$

$ab=0$のとき $\dps I_{12}=\int_a^b \cos x\,\d x=\sin b-\sin a$
$ab>0$のとき $t=x-\dfrac{ab}{x}$とおくと, $\d t=\cbk{1+\dfrac{ab}{x^2}}\,\d x,\,t=x-\dfrac{ab}{x}\Leftrightarrow x=\dfrac{t\pm\sqrt{t^2+4ab}}{2},\,1+\dfrac{ab}{x^2}=2-\dfrac{t}{x}$
$$ I_{12}=\int_{a-b}^{b-a} \frac{x}{2x-t}\cos t\,\d t $$

偶関数, 奇関数の積分の性質より
$$ I_{12}=\int_{a-b}^{b-a} \frac{t\pm\sqrt{t^2+4ab}}{\pm2\sqrt{t^2+4ab}}\cos t\,\d t =\frac{1}{2} \int_{a-b}^{b-a} \frac{\pm t\sqrt{t^2+4ab}+t^2+4ab}{t^2+4ab}\cos t\,\d t =\int_0^{b-a} \cos t\,\d t =\sin(b-a) $$
$$ \therefore I_{12}=\sin(b-a) $$

$$ I_{13}=\int \frac{(x\cos x-\sin x)'}{(x\cos x-\sin x)^2}\cdot\cbk{-\frac{x}{\sin x}}\,\d x =\frac{x}{(x\cos x-\sin x)\sin x}+\int \frac{\d x}{\sin^2 x} =\frac{x}{(x\cos x-\sin x)\sin x}-\frac{1}{\tan x}+C $$
$$ \therefore I_{13}=\frac{x\sin x+\cos x}{x\cos x-\sin x}+C $$

$y=x^{\frac{3}{2}}$とおくと, $\d y=\dfrac{3}{2}x^{\frac{1}{2}}\,\d x$
$$ I_{14}=\frac{2}{3}\int_0^1 y^{2n}\sqrt{1-y^2}\,\d y $$

$n=0$のとき, 円の積分になるから
$$ I_{14}=\frac{2}{3}\int_0^1 \sqrt{1-y^2}\,\d y=\frac{\pi}{6} $$

$n\geq1$のとき, $y=\sin t$とおくと, $\d y=\cos t\,\d t$
$$ I_{14}=\frac{2}{3}\int_0^{\frac{\pi}{2}} \sin^{2n} t\cos^2 t\,\d t =\frac{2}{3}\int_0^{\frac{\pi}{2}} (\sin^{2n} t-\sin^{2n+2} t)\,\d t =\frac{2}{3}\tbk{\frac{(2n-1)!!}{(2n)!!}-\frac{(2n+1)!!}{(2n+2)!!}}\frac{\pi}{2} $$
$$ I_{14}=\frac{2(2n-1)!!}{3(2n)!!}\cbk{1-\frac{2n+1}{2n+2}}\frac{\pi}{2} =\frac{(2n-1)!!}{3(2n+2)!!}\pi =\frac{(2n)!}{4^n n!^2(n+1)}\cdot\frac{\pi}{6} $$
$$ \therefore I_{14}=\frac{(2n)!}{4^n n!^2(n+1)}\cdot\frac{\pi}{6} $$

実数$\alpha$は$-\dfrac{\pi}{2}<\alpha<0,\,\tan\alpha=-\dfrac{4}{3}$を満たす. $t=\tan\theta$とおくと, $\d t=(t^2+1)\,\d x$
$$ I_{15}=\int_{\alpha}^0 \frac{(2\tan\theta+1)^3(\tan\theta-2)^3}{(3\tan\theta-1)^3(\tan\theta+3)^3}\,\d \theta =\int_{\alpha}^0 \frac{(2\sin^2\theta-3\sin\theta\cos\theta-2\cos^2\theta)^3}{(3\sin^2\theta+8\sin\theta\cos\theta-3\cos^2\theta)^3}\,\d \theta =\int_{\alpha}^0 \frac{(3\sin 2\theta+4\cos 2\theta)^3}{(4\sin 2\theta-3\cos 2\theta)^3}\,\d \theta =\int_{\alpha}^0 \tan^3(2\theta-\alpha)\,\d \theta $$
$$ I_{15}=\frac{1}{2}\int_{2\alpha}^0 \tan^3(\theta-\alpha)\,\d \theta =\frac{1}{2}\int_{2\alpha}^0 (\tbk{\tan(\theta-\alpha)}'-1)\tan(\theta-\alpha)\,\d \theta =\frac{1}{4}\bbk{\tan^2(\theta-\alpha)+2\log|\cos(\theta-\alpha)|}_{2\alpha}^0 $$
$$ \therefore I_{15}=0 $$

$$ I_{16}=\int_0^\frac{\pi}{2} \sqrt[3]{\frac{-\sin x+\sin^3 x+\cos x\sin x}{(1+\cos x)^2}}\,\d x =\int_0^\frac{\pi}{2} \sqrt[3]{\frac{\cos x\sin x(1-\cos x)}{(1+\cos x)^2}}\,\d x =\int_0^\frac{\pi}{2} \tan\frac{x}{2}\sqrt[3]{\cos x}\,\d x $$

$t^3=\cos x$とおくと, $\d x=-\dfrac{3t^2}{\sin x}\,\d t$
$$ I_{16}=\int_0^1 \tan\frac{x}{2}\cdot\frac{3t^2}{\sin x}\,\d t =\int_0^1 \frac{3t^3}{1+t^3}\,\d t =3-3\int_0^1 \frac{\d t}{1+t^3} =3-\int_0^1 \cbk{\frac{1}{t+1}+\sqrt{3}\cdot\frac{\cbk{(2t-1)/\sqrt{3}}'}{\cbk{(2t-1)/\sqrt{3}}^2+1}-\frac{1}{2}\cdot\frac{2t-1}{t^2-t+1}}\,\d t $$
$$ I_{16}=3-\bbk{\log|1+t|-\frac{1}{2}\log|t^2-t+1|+\sqrt{3}\arctan\cbk{(2t-1)/\sqrt{3}}}_0^1 =3-\log 2+\frac{\pi}{\sqrt{3}} $$

$x=\cos t$とおくと, $\d x=-\sin t\,\d t$
$$ I_{17}=\int_0^{\frac{\pi}{4}} \frac{(\cos^4 t-\cos^6 t)\sin^2 t}{32\cos^8 t-64\cos^6 t+40\cos^4 t-8\cos^2 t+1}\,\d t =\int_0^{\frac{\pi}{4}} \frac{(\cos^4 t-\cos^6 t)\sin^2 t}{1-8\cos^2 t(1-\cos^2 t)(2\cos^2 t-1)^2}\,\d t =\int_0^{\frac{\pi}{4}} \frac{\cos^4 t\sin^4 t}{1-8\cos^2 t\sin^2 t\cos^2 2t}\,\d t $$
$$ =\frac{1}{16}\int_0^{\frac{\pi}{4}} \frac{\sin^4 2t}{1-2\sin^2 2t\cos^2 2t}\,\d t =\frac{1}{16}\int_0^{\frac{\pi}{4}} \frac{\sin^4 2t}{\cos^4 2t+\sin^4 2t}\,\d t =\frac{1}{32}\int_0^{\frac{\pi}{2}} \frac{\sin^4 t}{\cos^4 t+\sin^4 t}\,\d t $$
King Propertyより, $s=\dfrac{\pi}{2}-t$とおく.
$$ I_{17}=\frac{1}{32}\int_0^{\frac{\pi}{2}} \frac{\cos^4 s}{\cos^4 s+\sin^4 s}\,\d s =\frac{1}{64}\int_0^{\frac{\pi}{2}} \d t =\frac{\pi}{128} $$

$$ I_{18}=\int_0^1 \frac{(x^2+1)'}{(x^2+1)^2}\cdot\frac{x\log(x+1)}{2}\,\d x =\bbk{-\frac{x \log(x+1)}{2(x^2+1)}}_0^1+\frac{1}{2}\int_0^1 \frac{1}{x^2+1}\cbk{\log(1+x)+\frac{x}{1+x}}\,\d x $$
$$ =-\frac{1}{4}\log 2+\frac{1}{2}\int_0^1 \frac{1}{(x+1)(x^2+1)}\,\d x+\frac{1}{2}\int_0^1 \frac{\log(1+x)}{x^2+1}\,\d x =-\frac{1}{4}\log 2+\frac{1}{4}\int_0^1 \cbk{\frac{x+1}{x^2+1}-\frac{1}{x+1}}\,\d x+\frac{1}{2}\int_0^1 \frac{\log(1+x)}{x^2+1}\,\d x $$
$$ =-\frac{1}{4}\log 2+\frac{1}{4}\bbk{\frac{1}{2}\log|x^2+1|-\log|x+1|+\arctan x}_0^1+\frac{1}{2}\int_0^1 \frac{\log(1+x)}{x^2+1}\,\d x =-\frac{3}{8}\log 2+\frac{\pi}{16}+\frac{1}{2}\int_0^1 \frac{\log(1+x)}{x^2+1}\,\d x $$

$\dfrac{\log(1+x)}{x^2+1}$の積分について$x=\tan t$とおくと, $\d x=(\tan^2 t+1)\,\d t$
$$ \int_0^1 \frac{\log(1+x)}{x^2+1}\,\d x=\int_0^{\frac{\pi}{4}} \log(1+\tan t)\,\d t $$

King Propertyより, $s=\dfrac{\pi}{4}-t$とおく.
$$ \int_0^{\frac{\pi}{4}} \log(1+\tan t)\,\d t =\int_0^{\frac{\pi}{4}} \log(1+\tan \cbk{\frac{\pi}{4}-s})\,\d s =\int_0^{\frac{\pi}{4}} \log(1+\frac{1-\tan s}{1+\tan s})\,\d s =\frac{\pi}{4}\log 2-\int_0^{\frac{\pi}{4}} \log(1+\tan s)\,\d s $$
$$ \therefore \int_0^{\frac{\pi}{4}}\log(1+\tan t)\,\d t=\frac{\pi}{8}\log 2\qquad \therefore I_{18}=\frac{\pi}{16}+\frac{\pi-6}{16}\log 2 $$

部分分数分解をする.
$$ \frac{1}{x^n(x^2-1)}=\frac{A}{x-1}+\frac{B}{x+1}+\sum_{k=1}^n \frac{C_k}{x^k} $$
$$ 1=Ax^n(x+1)+Bx^n(x-1)+(x^2-1)\sum_{k=1}^n C_k x^{n-k} $$
係数比較して
$$ A+B+C_1=0,\,A-B+C_2=0,\,C_n=-1,\,C_{n-1}=0,\,C_{m+2}-C_m=0\,(m=1,\,2,\,\cdots,\,n-2)\qquad\therefore A=-\frac{1}{2},\,B=\frac{(-1)^{n-1}}{2},\,C_m=-\frac{1+(-1)^{n+m}}{2} $$

$n\geq 1$で代入できるように調整して
$$ I_{19}=\int\frac{1}{x^n(x^2-1)}\,\d x=\int\cbk{\frac{A}{x-1}+\frac{B}{1+x}+\sum_{k=1}^n \frac{C_k}{x^k}}\,\d x=-A\log|1-x|+B\log|1+x|+C_1\log|x|+\frac{C_{n+1}}{n} x^{-n}-\sum_{k=1}^n \frac{C_{k+1}}{k} x^{-k}+C $$
$$ I_{19}=\frac{1}{2}\log|1-x|+\frac{(-1)^{n-1}}{2}\log|1+x|-\frac{1+(-1)^{n+1}}{2}\log|x|+\sum_{k=1}^n \frac{1+(-1)^{n+k-1}}{2k} x^{-k}+C $$

$y=x-\dfrac{1}{x}$とおくと, $\d y=\cbk{1+\dfrac{1}{x^2}}\,\d x$
$$ I_{20}=\int \frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,\d x =\int \frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\cdot\frac{x^2}{1+x^2}\,\d y =-\int \frac{x}{|x|}\cdot\frac{\d y}{(x-1/x)\sqrt{x^2+1/x^2}} =-\sgn x\int \frac{\d y}{y\sqrt{y^2+2}} $$

$t=\dfrac{\sqrt{2}}{y}$とおくと, $\d t=-\dfrac{\sqrt{2}}{y^2}\,\d y$
$$ I_{20}=\sgn x\int \frac{1}{y\sqrt{y^2+2}}\cdot\frac{y^2}{\sqrt{2}}\,\d t =\frac{\sgn x}{\sqrt{2}}\int \frac{|t|}{t}\cdot\frac{1}{\sqrt{1+t^2}}\,\d t =\frac{|t|}{\sqrt{2}t}\sgn x\arsinh t+C $$
$$ \therefore I_{20}=\frac{1}{\sqrt{2}}\sgn (x^2-1)\arsinh \cbk{\frac{\sqrt{2}x}{x^2-1}}+C $$

King Propertyで, $y=-x$とおく.
$$ I_{21}=\int_{-\pi}^\pi \frac{\sin^{2n} y}{(1+e^{-y})\sqrt{1+\cos y}}\,\d y =\frac{1}{2} \int_{-\pi}^\pi \frac{\sin^{2n} x}{\sqrt{1+\cos x}}\,\d x =\int_0^\pi \frac{\sin^{2n} x}{\sqrt{1+\cos x}}\,\d x =\frac{2^{2n}}{\sqrt{2}}\int_0^\pi \sin^{2n} \frac{x}{2}\cos^{2n-1} \frac{x}{2}\,\d x $$
$$ I_{21}=2^{2n}\sqrt{2}\int_0^{\frac{\pi}{2}} \sin^{2n} x\cos^{2n-1} x\,\d x $$

• ベータ関数を使う場合
  ジャンドルの倍数公式$\Gamma(x)\Gamma(x+1/2)=2^{1-2x}\sqrt{\pi}\,\Gamma(2x)$で変形する.
  $$ I_{21}=2^{n-1}\sqrt{2}\cdot\frac{\Gamma(n+1/2)\Gamma(n)}{\Gamma(2n+1/2)} =2^{n-1}\sqrt{2}\cdot\frac{2^{1-2n}\sqrt{\pi}\,\Gamma(2n)}{\Gamma(2n+1/2)} =2^{n-1}\sqrt{2}\cdot\frac{2^{1-2n}\sqrt{\pi}\,\Gamma(2n)^2}{\Gamma(2n)\Gamma(2n+1/2)} =2^{n-1}\sqrt{2}\cdot\frac{2^{1-2n}\sqrt{\pi}\,\Gamma(2n)^2}{2^{1-4n}\sqrt{\pi}\,\Gamma(4n)} $$
  $$ \therefore I_{21}=\frac{2^{3n-1}(2n-1)!^2\sqrt{2}}{(4n-1)!} $$

• ベータ関数を使わない場合
  $t=\sin x$とおくと, $\d t=\cos x\,\d x$
  $$ I_{21}=2^{2n}\sqrt{2}\int_0^1 t^{2n}(1-t^2)^{n-1}\,\d t $$

$\dps F(2n,\,n-1)=\int_0^1 t^{2n}(1-t^2)^{n-1}\,\d t$として, 漸化式を導く.
$$ F(2n,\,n-1)=\bbk{\frac{1}{2n+1}t^{2n+1}(1-t^2)^{n-1}}_0^1+\frac{2(n-1)}{2n+1}\int_0^1 t^{2n+2}(1-t^2)^{n-2}\,\d t =\frac{2(n-1)}{2n+1}F(2(n+1),\,n-2) $$

$m$を自然数として, $m$回部分積分をした項まで計算する.
$$ F(2n,\,n-1) =\frac{2^2(n-1)(n-2)}{(2n+1)(2n+3)}F(2(n+2),\,n-3) =\frac{2^m(n-1)(n-2)\cdots(n-m)}{(2n+1)(2n+3)\cdots(2n+2m-1)}F(2(n+m),\,n-m-1) $$

$m=n-1$を代入するが, $n\geq 1$なので$m\geq 0$で対応するようにすると
$$ F(2n,\,n-1) =\frac{2^m (n-1)!(2n-1)!!}{(n-m-1)!(2n+2m-1)!!}F(2(n+m),\,n-m-1) =\frac{2^{n-1} (n-1)!(2n-1)!!}{0!(4n-3)!!}F(4n-2,\,0) $$

二重階乗を計算すると
$$ F(2n,\,n-1)=\frac{2^{n-1} (n-1)!(2n-1)!!}{0!(4n-3)!!}\int_0^1 t^{4n-2}(1-t^2)^0\,\d t =\frac{2^{n-1} (n-1)!(2n-1)!!}{(4n-1)!!} =\frac{2^{n-1} (2n-1)!^2}{(4n-1)!} $$
$$ \therefore I_{21}=\frac{2^{3n-1}(2n-1)!^2\sqrt{2}}{(4n-1)!} $$

二重根号を外す. $\dps \sqrt{A}+\sqrt{B}=\sqrt{\frac{x^2+1+\sqrt{x^4-x^2+1}}{x^4-x^2+1}}$を満たすような$A,\,B$を求める.
$\dps A+B+2\sqrt{AB}=\frac{x^2+1+\sqrt{x^4-x^2+1}}{x^4-x^2+1}$より, $\dps A+B=\frac{x^2+1}{x^4-x^2+1},\,AB=\frac{1}{4x^4-4x^2+4}$
$$ (A-B)^2=(A+B)^2-4AB=\cbk{\frac{x^2+1}{x^4-x^2+1}}^2-\frac{1}{x^4-x^2+1}=\frac{3x^2}{(x^4-x^2+1)^2} $$
$$ A,\,B=\frac{A+B\pm|A-B|}{2} =\frac{A+B\pm\sqrt{(A-B)^2}}{2} =\frac{x^2\pm\sqrt{3} x+1}{2(x^4-x^2+1)} =\frac{1}{2(x^2\mp\sqrt{3} x+1)} $$

$$ I_{22}=\frac{1}{\sqrt{2}}\int_0^1 \cbk{\frac{1}{\sqrt{x^2+\sqrt{3} x+1}}+\frac{1}{\sqrt{x^2-\sqrt{3} x+1}}\,}\,\d x =\frac{1}{\sqrt{2}}\int_0^1 \cbk{\frac{(2x+\sqrt{3}\,)'}{\sqrt{(2x+\sqrt{3}\,)^2+1}}+\frac{(2x-\sqrt{3}\,)'}{\sqrt{(2x-\sqrt{3}\,)^2+1}}\,}\,\d x $$
$$ I_{22}=\frac{1}{\sqrt{2}}\bbk{\arsinh(2x+\sqrt{3}\,)+\arsinh(2x-\sqrt{3}\,)}_0^1 =\frac{1}{\sqrt{2}}\bbk{\log\cbk{2x+\sqrt{3}\,+\sqrt{(2x+\sqrt{3})^2+1\,}\,}+\log\cbk{2x-\sqrt{3}\,+\sqrt{(2x-\sqrt{3})^2+1\,}\,}}_0^1 $$
$$ I_{22}=\sqrt{2}\log(\sqrt{2}+\sqrt{3}) $$

$y=\dfrac{1}{x}$とおくと, $\d y=-\dfrac{1}{x^2}\,\d x$
$$ I=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{1+y^2}f\left(\frac{y^2}{y^2+\sqrt{2}}\right)\,\d y =\frac{1}{2}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{1+x^2}\left\{f\left(\frac{x^2}{x^2+\sqrt{2}}\right)+f\left(\frac{1}{1+\sqrt{2}x^2}\right)\right\}\,\d x $$
ここで$\tan\left\{f\left(\frac{x^2}{x^2+\sqrt{2}}\right)+f\left(\frac{1}{1+\sqrt{2}x^2}\right)\right\}=\dfrac{\frac{x^2}{x^2+\sqrt{2}}+\frac{1}{1+\sqrt{2}x^2}}{1-\frac{x^2}{x^2+\sqrt{2}}\cdot\frac{1}{1+\sqrt{2}x^2}}=1$なので, $f\left(\frac{x^2}{x^2+\sqrt{2}}\right)+f\left(\frac{1}{1+\sqrt{2}x^2}\right)=\dfrac{\pi}{4}$となる.
$$ I=\frac{\pi}{8}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{1+x^2}\,\d x =\frac{\pi}{8}\bbk{\arctan x}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} =\frac{\pi^2}{48} $$

$2t=\cos2x$とおくと, $\d t=-\sin2x\,\d x$
$$ I=\int_0^{\frac{\pi}{2}} \cos x\sin x\cbk{e^{\frac{1+\cos2x}{2}}+e^{\frac{1-\cos2x}{2}}}(\cos^4 x+\sin^4 x)\,\d x =\frac{\sqrt{e}}{2}\int_{-\frac{1}{2}}^{\frac{1}{2}} (1+4t^2)\cosh t\,\d t $$

部分積分$\dps\int t^2\cosh t\,\d t=t^2\sinh t-\int 2t\sinh t\,\d t=t^2\sinh t-2t\cosh t+2\int \cosh t\,\d t=t^2\sinh t-2t\cosh t+2\sinh t$より
$$ I=\frac{\sqrt{e}}{2}\bbk{(4t^2+9)\sinh t-8t\cosh t}_{-\frac{1}{2}}^{\frac{1}{2}} =3e-7 $$

King Propertyより2積分を足して, $1+e^x$を消す.

$$ I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\d x}{(\cos^8 x+\sin^8 x)(1+e^{-x})} =\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\d x}{\cos^8 x+\sin^8 x} =\int_0^{\frac{\pi}{2}} \frac{\d x}{\cos^8 x+\sin^8 x} $$

$\dps\cos^8 \cbk{\frac{\pi}{4}+x}+\sin^8 \cbk{\frac{\pi}{4}+x}=\cos^8 \cbk{\frac{\pi}{4}-x}+\sin^8 \cbk{\frac{\pi}{4}-x}$が成り立つので, $x=\dfrac{\pi}{4}$で対称性が言えて
$$ I=2\int_0^{\frac{\pi}{4}} \frac{\d x}{\cos^8 x+\sin^8 x} =2^5\int_0^{\frac{\pi}{4}} \frac{\d x}{(1+\cos 2x)^4+(1-\cos 2x)^4} =2^4\int_0^{\frac{\pi}{4}} \frac{\d x}{\cos^4 2x+6\cos^2 2x+1} =2^3\int_0^{\frac{\pi}{2}} \frac{\d x}{\cos^4 x+6\cos^2 x+1} $$

部分分数分解をして
$$ =\sqrt{2}\int_0^{\frac{\pi}{2}} \cbk{\frac{1}{\cos^2 x+3-2\sqrt{2}}-\frac{1}{\cos^2 x+3+2\sqrt{2}}}\,\d x =\sqrt{2}\int_0^{\frac{\pi}{2}} \cbk{\frac{1}{4-2\sqrt{2}-\sin^2 x}-\frac{1}{4+2\sqrt{2}-\sin^2 x}}\,\d x $$

式が複雑になるので, $a=\sqrt{4-2\sqrt{2}},\,b=\sqrt{4+2\sqrt{2}}$とおき, 前半と後半に分けて積分する.
$$ \text{前半}=\sqrt{2}\int_0^{\frac{\pi}{2}} \frac{\d x}{4-2\sqrt{2}-\sin^2 x}=\frac{1}{\sqrt{2}a}\int_0^{\frac{\pi}{2}} \cbk{\frac{1}{a-\sin x}+\frac{1}{a+\sin x}}\,\d x $$
$$ \text{後半}=\sqrt{2}\int_0^{\frac{\pi}{2}} \frac{\d x}{4+2\sqrt{2}-\sin^2 x}=\frac{1}{\sqrt{2}b}\int_0^{\frac{\pi}{2}} \cbk{\frac{1}{b-\sin x}+\frac{1}{b+\sin x}}\,\d x $$

$t=\tan\dfrac{x}{2}$とおくと, $\d t=\dfrac{t^2+1}{2}\,\d x$
$$ \text{前半}=\frac{\sqrt{2}}{a}\int_0^1 \cbk{\frac{1}{at^2-2t+a}+\frac{1}{at^2+2t+a}}\,\d t =\frac{\sqrt{2}}{a^2}\int_0^1 \cbk{\frac{1}{(t-1/a)^2+1-1/a^2}+\frac{1}{(t+1/a)^2+1-1/a^2}}\,\d t $$
$$ \text{後半}=\frac{\sqrt{2}}{b}\int_0^1 \cbk{\frac{1}{bt^2-2t+b}+\frac{1}{bt^2+2t+b}}\,\d t =\frac{\sqrt{2}}{b^2}\int_0^1 \cbk{\frac{1}{(t-1/b)^2+1-1/b^2}+\frac{1}{(t+1/b)^2+1-1/b^2}}\,\d t $$

$\dps\int \frac{\d x}{x^2+c^2}=\frac{1}{c}\arctan\frac{x}{c}+C,\,\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$より
$$ \text{前半}=\frac{\sqrt{2}}{a\sqrt{a^2-1}}\bbk{\arctan\frac{at-1}{\sqrt{a^2-1}}+\arctan\frac{at+1}{\sqrt{a^2-1}}}_0^1 =\frac{2\sqrt{2}}{a\sqrt{a^2-1}}\cbk{\arctan\frac{a-1}{\sqrt{a^2-1}}+\arctan\frac{a+1}{\sqrt{a^2-1}}} =\frac{\sqrt{2}}{2a\sqrt{a^2-1}}\pi $$
$$ \text{前半}=\frac{\sqrt{2}}{b\sqrt{b^2-1}}\bbk{\arctan\frac{bt-1}{\sqrt{b^2-1}}+\arctan\frac{bt+1}{\sqrt{b^2-1}}}_0^1 =\frac{2\sqrt{2}}{b\sqrt{b^2-1}}\cbk{\arctan\frac{b-1}{\sqrt{b^2-1}}+\arctan\frac{b+1}{\sqrt{b^2-1}}} =\frac{\sqrt{2}}{2b\sqrt{b^2-1}}\pi $$

$ab=2\sqrt{2},\,\sqrt{a^2-1}=\sqrt{2}-1,\,\sqrt{b^2-1}=\sqrt{2}+1$より
2積分を合わせて, $I$を次のようにa, bを使って表せる.
$$ I=\frac{\sqrt{2}}{2a\sqrt{a^2-1}}\pi-\frac{\sqrt{2}}{2b\sqrt{b^2-1}}\pi =\frac{(1+\sqrt{2})b-(\sqrt{2}-1)a}{4}\pi $$

$a^2+b^2=8,\,ab=2\sqrt{2}$より, $(b-a)^2=a^2+b^2-2ab=8-4\sqrt{2}=2a^2\qquad\therefore b=(1+\sqrt{2})a$
$$ \therefore I=\frac{(1+\sqrt{2})^2-(\sqrt{2}-1)}{4}a\pi=\frac{(4+\sqrt{2})\sqrt{4- 2\sqrt{2}}}{4}\pi $$

半角の公式より
$$ (1+\cos x)\cbk{1+\cos \frac{x}{2}}\cdots\cbk{1+\cos \frac{x}{2^{n-1}}} =2^n \cos^2\frac{x}{2}\cos^2\frac{x}{4}\cdots\cos^2\frac{x}{2^n} =\frac{\sin^2 x}{2^n\sin^2\frac{x}{2^n}}\quad\cbk{\sin^2\frac{x}{2^n}\neq0} $$
$x=0,\,2^n\pi$のとき, $\sin^2\dfrac{x}{2^n}=0$である. $\dps \lim_{x\to0}\frac{\sin^2 x}{2^n\sin^2\frac{x}{2^n}}=2^n,\,\lim_{x\to 2^n\pi}\frac{\sin^2 x}{2^n\sin^2\frac{x}{2^n}}=2^n$より
$m$を整数として$\dps (1+\cos x)\cbk{1+\cos \frac{x}{2}}\cdots\cbk{1+\cos \frac{x}{2^{n-1}}}=\begin{cases} \dfrac{\sin^2 x}{2^n\sin^2\frac{x}{2^n}} && \text{if $x\neq 2^n m\pi$,} \\ \quad 2^n && \text{if $x=2^n m\pi$.} \end{cases}$とすれば, 連続となる.

$x=2^n y$とおくと, $\dps\d x=2^n\d y,\,f(x)=\begin{cases} \dfrac{\sin^2 2^n x}{\sin^2 x} && \text{if $x\neq m\pi$,} \\ \quad 2^n && \text{if $x=m\pi$.} \end{cases}$として

$$ I=\int_0^\pi f(y)\,\d y $$
ここで$\cos$の総和を考えると, $\dps 2\sum_{k=0}^{2^{n-1}-1} \cos(2k+1)x=\frac{\sin 2^n x}{\sin x}$なので
$$ I=\int_0^\pi \cbk{2\sum_{k=0}^{2^{n-1}-1} \cos(2k+1)x}^2\,\d x =4\sum_{k=0}^{2^{n-1}-1}\sum_{l=0}^{2^{n-1}-1}\int_0^\pi \cos(2k+1)x \cos(2l+1)x\,\d x $$
$k\neq l$のとき, $\dps \int_0^\pi \cos(2k+1)x \cos(2l+1)x\,\d x=\frac{1}{2}\bbk{\frac{\sin(2k+2l+2)x}{2k+2l+2}+\frac{\sin(2k-2l)x}{2k-2l}}_0^\pi=0$
$k=l$のとき, $\dps \int_0^\pi \cos(2k+1)x \cos(2l+1)x\,\d x=\bbk{\frac{1+\cos(4k+2)x}{2}}_0^\pi=\frac{\pi}{2}$
$$ \therefore I=4\sum_{k=0}^{2^{n-1}-1} \frac{\pi}{2}=2^n \pi $$

式$(\sqrt{A}+\sqrt{B})^2=A+B+2\sqrt{AB}$より
$$ \sqrt{\sqrt{x}+\sqrt{x-x^2}}=\frac{\sqrt[4]{x}}{\sqrt{2}}\cbk{\sqrt{1-\sqrt{x}}+\sqrt{1+\sqrt{x}}} $$
と二重根号を変形できる.

式$(A\sqrt[4]{x}+B\sqrt[4]{x^3})^2=2ABx+(A^2+B^2x)\sqrt{x}$より
$$ \sqrt{-2x+(1+x)\sqrt{x}}=\sqrt[4]{x}-\sqrt[4]{x^3} $$
と二重根号を変形できる.

$$ I=\int_0^1 \cbk{1+\sqrt[4]{x}-\sqrt[4]{x^3}}\tbk{1+\frac{\sqrt[4]{x}}{\sqrt{2}}\cbk{\sqrt{1-\sqrt{x}}+\sqrt{1+\sqrt{x}}}}\,\d x $$
展開して個別に積分をする.

$1+\sqrt[4]{x}-\sqrt[4]{x^3}$の積分

$$ J_1=\int_0^1 (1+\sqrt[4]{x}-\sqrt[4]{x^3})\,\d x =\bbk{x+\frac{4}{5}x^{\frac{5}{4}}-\frac{4}{7}x^{\frac{7}{4}}}_0^1 =\frac{43}{35} $$

$\sqrt[4]{x}\sqrt{1-\sqrt{x}}$の積分

$$ J_2=\int_0^1 \sqrt[4]{x}\sqrt{1-\sqrt{x}}\,\d x $$
$x=\sin^4 t$とおく.
$$ J_2=\int_0^{\frac{\pi}{2}} \sin t\sqrt{1-\sin^2 t}\cdot4\cos t\sin^3 t\,\d t =4\int_0^{\frac{\pi}{2}} (\sin^4 t-\sin^6 t)\,\d t =4\cbk{\frac{3!!}{4!!}-\frac{5!!}{6!!}}\frac{\pi}{2} =\frac{\pi}{8} $$

$\sqrt[4]{x}\sqrt{1+\sqrt{x}}$の積分

$$ J_3=\int_0^1 \sqrt[4]{x}\sqrt{1+\sqrt{x}}\,\d x $$
$x=\sinh^4 t$とおく.
$$ J_3=\int_0^{\log(1+\sqrt{2})} \sinh t\sqrt{1+\sinh^2 t}\cdot 4\cosh t\sinh^3 t\,\d t =4\int_0^{\log(1+\sqrt{2})} (\sinh^4 t+\sinh^6 t)\,\d t $$
ここで$\dps \int \sinh^{2n} t\,\d t=-\frac{2n-1}{2n}\int \sinh^{2n-2} t\,\d t+\frac{1}{2n}\cosh t\sinh^{2n-1} t$が成り立つので
$$ J_3=4\int_0^{\log(1+\sqrt{2})} (\sinh^4 t+\sinh^6 t)\,\d t =\frac{2}{3}\bbk{\cosh t\sinh^5 t}_0^{\log(1+\sqrt{2})}+\frac{2}{3}\int_0^{\log(1+\sqrt{2})} \sinh^4 t\,\d t $$
$$ =\bbk{\frac{2}{3}\cosh t\sinh^5 t+\frac{1}{6}\cosh t\sinh^3 t}_0^{\log(1+\sqrt{2})}-\frac{1}{2}\int_0^{\log(1+\sqrt{2})} \sinh^2 t\,\d t =\bbk{\frac{2}{3}\cosh t\sinh^5 t+\frac{1}{6}\cosh t\sinh^3 t-\frac{1}{4}\cosh t\sinh t+\frac{1}{4}t}_0^{\log(1+\sqrt{2})} $$
$$ \therefore J_3=\frac{7\sqrt{2}}{12}+\frac{1}{4}\log(1+\sqrt{2}) $$

$(\sqrt{x}-x)\sqrt{1-\sqrt{x}}$の積分

$$ J_4=\int_0^1 (\sqrt{x}-x)\sqrt{1-\sqrt{x}}\,\d x $$

$t=\sqrt{1-\sqrt{x}}$とおく.
$$ J_4=4\int_0^1 t^4 (1-t^2)^2\,\d t =4\bbk{\frac{1}{5}t^5-\frac{2}{7}t^7+\frac{1}{9}t^9}_0^1 =\frac{32}{315} $$

$(\sqrt{x}-x)\sqrt{1+\sqrt{x}}$の積分

$$ J_5=\int_0^1 (\sqrt{x}-x)\sqrt{1+\sqrt{x}}\,\d x $$

$t=\sqrt{1+\sqrt{x}}$とおく.
$$ J_5=4\int_1^{\sqrt{2}} t(t^2-1)^2(2-t^2)\,\d t =4\bbk{-\frac{1}{9}t^9+\frac{4}{7}t^7-t^5+\frac{2}{3}t^3}_1^{\sqrt{2}} =\frac{32}{63}(\sqrt{2}-1) $$

$$ I=J_1+\frac{1}{\sqrt{2}}(J_2+J_3+J_4+J_5) =\frac{2923}{1260}-\frac{64\sqrt{2}}{315}+\frac{\sqrt{2}}{16}\pi+\frac{\sqrt{2}}{8}\log(1+\sqrt{2}) $$