あなたも√sinxの積分を求めませんか？

\begin{eqnarray}B(x,y)&=&\int^1_0t^{x-1}(1-t)^{y-1}dt\qquad t=\sin^2\theta,dt=2\sin\theta\cos\theta \\[7pt]&=&\int^{\frac{\pi}{2}}_0\sin^{2x-2}\theta(1-\sin^2\theta)^{y-1}\cdot2\sin\theta\cos\theta d\theta\\[7pt]&=&2\int^{\frac{\pi}{2}}_0\sin^{2x-1}\theta\cos^{2y-1}\theta d\theta\end{eqnarray}

\begin{eqnarray}\int^{\frac{\pi}{2}}_0\!\!\sqrt{\sin x}dx&=&\int_{\frac{\pi}{2}}^0\!\! -\sqrt{\sin \left(\frac{\pi}{2}-t\right)}dt\qquad x=\frac{\pi}{2}-t,dx=-dt\\[7pt]&=&\int^{\frac{\pi}{2}}_0\!\!\sqrt{\cos t}dt\end{eqnarray}

\begin{eqnarray}\int^{\frac{\pi}{2}}_0\!\!\sqrt{\sin x}dx&=&\int^{\frac{\pi}{2}}_0\!\!\sin^{\frac{1}{2}} xdx\\[7pt]&=&\int^{\frac{\pi}{2}}_0\!\!\sin^{\frac{1}{2}} x\cdot\cos^0x dx\\[7pt]&=&\int^{\frac{\pi}{2}}_0\!\!\sin^{2\cdot\frac{3}{4}-1} x\cdot\cos^{2\cdot\frac{1}{2}-1}x dx\\[7pt]&=&\frac{1}{2}B\left(\frac{3}{4},\frac{1}{2}\right)\\[7pt]&=&\frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{5}{4}\right)}\\[7pt]&=&\frac{\sqrt{\pi}~\Gamma\left(\frac{3}{4}\right)}{2\Gamma\left(\frac{5}{4}\right)}\\[7pt]&=&\sqrt{\frac{2}{\pi}}\Gamma\left(\frac{3}{4}\right)^2\end{eqnarray}

\begin{eqnarray}\int^{\frac{\pi}{2}}_0\!\!\sqrt{\tan x}dx&=&\int^{\frac{\pi}{2}}_0\!\!\frac{\sqrt{\sin x}}{\sqrt{\cos x}}dx\\[7pt]&=&\int^{\frac{\pi}{2}}_0\!\!\sin^{\frac{1}{2}} x\cdot\cos^{-\frac{1}{2}}x dx \\[7pt]&=&\int^{\frac{\pi}{2}}_0\!\!\sin^{2\cdot\frac{3}{4}-1} x\cdot\cos^{2\cdot\frac{1}{4}-1}x dx\\[7pt]&=&\frac{1}{2}B\left(\frac{3}{4},\frac{1}{4}\right)\\[7pt]&=&\frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(1\right)}\\[7pt]&=&\frac{1}{2}\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{4}\right)\\[7pt]&=&\frac{1}{2}\Gamma\left(1-\frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)\\[7pt]&=&\frac{\pi}{2\sin{\frac{\pi}{4}}}\qquad相反公式を利用\\[7pt]&=&\frac{\pi}{\sqrt{2}} \end{eqnarray}

\begin{eqnarray}\int_0^x\sqrt{\cos t}dt&=&\int_0^x\sqrt{1-2\cdot\frac{1-\cos t}{2}}dt\\[7pt]&=&\int^x_0\sqrt{1-2\sin^2\frac{t}{2}}dt\\[7pt]&=&2\int^{\frac{x}{2}}_0\sqrt{1-2\sin^2\theta}d\theta\qquad t=2\theta,dt=2d\theta\end{eqnarray}

\begin{eqnarray}2\int^{\frac{x}{2}}_0\sqrt{1-2\sin^2\theta}d\theta&=&2\int^{\frac{x}{2}}_0\sqrt{1-\sqrt{2}^2\sin^2\theta}d\theta\\[7pt]&=&2E(\sqrt{2},\frac{x}{2})\end{eqnarray}

二次式 $ax^2 + bx + c$ を平方完成すると、
\begin{align*} ax^2 + bx + c &= a \left( x^2 + \frac{b}{a} x + \frac{c}{a} \right) \\ &= a \left( \left(x + \frac{b}{2a} \right)^2 - \frac{D}{4a^2} \right) \end{align*}
と変形できる。ここで$D=b^2-4ac$である。
$u = x + \frac{b}{2a}$と置くと、積分は次の形になる
$$I = \int \frac{dx}{a \left( u^2 - \frac{D}{4a^2} \right)}$$
ここで、$k^2 = -\frac{D}{4a^2}$ とおくと、
$$I = \frac{1}{a} \int \frac{du}{u^2 + k^2}$$
ここで以下の公式を用いる。
$$\int \frac{dx}{x^2 + {\alpha}^2} = \frac{1}{\alpha} \arctan \left( \frac{x}{\alpha} \right) + C$$
これを適用すると、
$$I = \frac{1}{a} \cdot \frac{1}{k} \arctan \left( \frac{u}{k} \right) + C$$
もとの変数 $x$ に戻すと
$$I = \frac{1}{a}\cdot\frac{2a}{\sqrt{-D}} \arctan \left( \frac{2a(x + \frac{b}{2a})}{\sqrt{-D}} \right) + C=\frac{2}{\sqrt{4ac-b^2}}\arctan\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C$$

$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$
に$A=\arctan\alpha,B=\arctan\beta$を代入すると、
$$\tan(\arctan\alpha+\arctan\beta)=\frac{\alpha+\beta}{1-\alpha\beta} $$
両辺$\arctan$をとると$$\arctan(α)+\arctan(β)=\arctan\left(\frac{α+β}{1-αβ}\right)$$

\begin{eqnarray}I=\int{\sqrt{\tan x}dx}&=&\int{ \frac{2t^2}{1+t^4}dt}\qquad t=\sqrt{\tan x},dx=\frac{2t}{1+t^4}dt\\[7pt]&=&\frac{1}{\sqrt{2}}\int\left(\frac{t}{t^2-\sqrt{2}t+1}-\frac{t}{t^2+\sqrt{2}t+1}\right)dt\end{eqnarray}
このように部分分数分解によって2つの積分に分けることができた。
左側の積分を$I_1$、右側の積分を$I_2$とすると、
$$\sqrt{2}I=I_1-I_2$$
となる。それぞれ計算していく。
\begin{eqnarray}I_1=\int\frac{t}{t^2-\sqrt{2}t+1}dt&=&\frac{1}{2}\left(\int\frac{2t-\sqrt{2}}{t^2-\sqrt{2}t+1}dt+\int\frac{\sqrt{2}}{t^2-\sqrt{2}t+1}dt\right)\\[7pt]&=&\frac{1}{2}\left(\int\frac{(t^2-\sqrt{2}t+1)'}{t^2-\sqrt{2}t+1}dt+\sqrt{2}\int\frac{1}{t^2-\sqrt{2}t+1}dt\right)\\[7pt]&=&\frac{1}{2}\log\left|t^2-\sqrt{2}t+1\right|+\arctan\left(\sqrt{2}t-1\right)\end{eqnarray}
\begin{eqnarray}I_2=\int\frac{t}{t^2+\sqrt{2}t+1}dt&=&\frac{1}{2}\left(\int\frac{2t+\sqrt{2}}{t^2+\sqrt{2}t+1}dt-\int\frac{\sqrt{2}}{t^2+\sqrt{2}t+1}dt\right)\\[7pt]&=&\frac{1}{2}\left(\int\frac{(t^2+\sqrt{2}t+1)'}{t^2+\sqrt{2}t+1}dt-\sqrt{2}\int\frac{1}{t^2+\sqrt{2}t+1}dt\right)\\[7pt]&=&\frac{1}{2}\log\left|t^2+\sqrt{2}t+1\right|+\arctan\left(\sqrt{2}t+1\right)\end{eqnarray}
よって、
\begin{eqnarray}I=\frac{1}{\sqrt{2}}(I_1-I_2)&=&\frac{1}{\sqrt{2}}\left(\left(\frac{1}{2}\log\left|t^2-\sqrt{2}t+1\right|+\arctan\left(\sqrt{2}t-1\right)\right)-\left(\frac{1}{2}\log\left|t^2+\sqrt{2}t+1\right|+\arctan\left(\sqrt{2}t+1\right)\right)\right)\\[7pt]&=&\frac{1}{2\sqrt{2}} \log \left| \frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1} \right|+\frac{1}{\sqrt{2}} \left(\arctan(\sqrt{2}t-1)+ \arctan(\sqrt{2}t+1)\right)\\[7pt]&=&\frac{1}{2\sqrt{2}} \log \left| \frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1} \right|+\frac{1}{\sqrt{2}} \left(\arctan(\frac{\sqrt{2}t}{1-t^2})\right)\\[7pt]&=&\frac{1}{2\sqrt{2}} \log \left| \frac{\tan(x)-\sqrt{2\tan(x)}+1}{\tan(x)+\sqrt{2\tan(x)}+1} \right|+\frac{1}{\sqrt{2}}  \arctan \left( \frac{\sqrt{2\tan(x)}}{1-\tan(x)} \right)\end{eqnarray}