sinの二乗の不定積分

sinの二乗の不定積分

$$ \begin{eqnarray} \int \sin^2{\left(\theta\right)}\mathrm{d}\theta &=&\int \frac{1}{2}-\frac{1}{2}\cos{\left(2\theta\right)}\mathrm{d}\theta \\&&\;\cdots\;\cos{\left(2\theta\right)}=\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)}=\left(1-\sin^2{\left(\theta\right)}\right)-\sin^2{\left(\theta\right)}=1-2\sin^2{\left(\theta\right)} \\&&\;\cdots\;\sin^2{\left(\theta\right)}=\frac{1}{2}-\frac{1}{2}\cos{\left(2\theta\right)} \\&=&\frac{1}{2}\int \mathrm{d}\theta-\frac{1}{2}\int\cos{\left(2\theta\right)}\mathrm{d}\theta \\&=&\frac{1}{2}\theta-\frac{1}{2}\int\cos{\left(2\theta\right)}\mathrm{d}\theta \\&=&\frac{1}{2}\theta-\frac{1}{2}\int\cos{\left(\phi\right)}\frac{1}{2}\mathrm{d}\phi \;\cdots\;\phi=2\theta, \frac{\mathrm{d}\phi}{\mathrm{d}\theta}=2,\mathrm{d}\theta=\frac{1}{2}\mathrm{d}\phi \\&=&\frac{1}{2}\theta-\frac{1}{4}\int\cos{\left(\phi\right)}\mathrm{d}\phi\;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{2}\theta-\frac{1}{2}\left(\frac{1}{2}\sin{\left(\phi\right)}\right) \;\cdots\;\int \cos{\left(x\right)}\mathrm{d}x= \sin{\left(x\right)}+C\;(C:積分定数) \\&=&\frac{1}{2}\theta-\frac{1}{4}\sin{\left(\phi\right)} \\&=&\frac{1}{2}\theta-\frac{1}{4}\sin{\left(2\theta\right)} \\&=&\frac{1}{2}\theta-\frac{1}{2}\sin{\left(\theta\right)}\cos{\left(\theta\right)} \\&=&\frac{1}{2}\left\{\theta-\sin{\left(\theta\right)}\cos{\left(\theta\right)}\right\}+C\;\cdots\;C:積分定数 \end{eqnarray} $$