
$\sin$ に置きかえる1
$\displaystyle\int_0^1 \sqrt{1-x^2}\space dx$

$x=\sin t$ とおくと
$dx=\cos t\space dt$
$x\space 0\rightarrow 1$
$\displaystyle t\space 0\rightarrow\frac{\pi}{2}$
$\displaystyle=\int_0^{\frac{\pi}{2}}\sqrt{1-\sin^2 t}\cdot\cos t\space dt$
$\displaystyle=\int_0^{\frac{\pi}{2}}\cos t\cdot\cos t\space dt$
$\displaystyle=\int_0^{\frac{\pi}{2}}\cos^2 t\space dt$

$\displaystyle=\int_0^{\frac{\pi}{2}}\frac{1+\cos 2t}{2}\space dt$
$\displaystyle=\frac{1}{2}\int_0^{\frac{\pi}{2}}1+\cos 2t\space dt$
$\displaystyle=\frac{1}{2}\left[t+\frac{1}{2}\sin 2t\right]_0^\frac{\pi}{2}$
$\displaystyle=\frac{1}{2}\left\{\left(\frac{\pi}{2}+\frac{1}{2}\cdot0\right)-(0+0)\right\}$
$\displaystyle=\frac{\pi}{4}$(答え)
$\sin$ に置きかえる2
$\displaystyle\int_0^2 \sqrt{4-x^2}\space dx$




実際に計算してみると何をしたいのかが分かります。
$x=2\sin t$ とおくと
$dx=2\cos t\space dt$
$x\space 0\rightarrow 2$
$\displaystyle t\space 0\rightarrow\frac{\pi}{2}$
$\displaystyle=\int_0^{\frac{\pi}{2}} \sqrt{4-4\sin^2 t}\cdot 2\cos t\space dt$
$\displaystyle=\int_0^{\frac{\pi}{2}} 2\sqrt{1-\sin^2 t}\cdot 2\cos t\space dt$
$\displaystyle=\int_0^{\frac{\pi}{2}} 2\cos t\cdot 2\cos t\space dt$
だから $1-\sin^2 x=\cos^2 x$



$\displaystyle=4\int_0^{\frac{\pi}{2}} \cos^2 t\space dt$
$\displaystyle=4\int_0^{\frac{\pi}{2}} \frac{1+\cos 2t}{2}\space dt$
$\displaystyle=2\int_0^{\frac{\pi}{2}} 1+\cos 2t\space dt$
$\displaystyle=2\left[t+\frac{1}{2}\sin 2t\right]_0^{\frac{\pi}{2}}$
$\displaystyle=2\left\{\left(\frac{\pi}{2}+0\right)-\left(0+0\right)\right\}$
$=\pi$(答え)
$\sin$ に置きかえる3
$\displaystyle\int_0^{\sqrt{3}} \sqrt{3-x^2}\space dx$



$x=\sqrt{3}\sin t$ とおくと
$dx=\sqrt{3}\cos t\space dt$
$x\space 0\rightarrow \sqrt{3}$
$\displaystyle t\space 0\rightarrow\frac{\pi}{2}$
$\displaystyle =\int_0^{\frac{\pi}{2}} \sqrt{3-3\sin^2 t}\cdot\sqrt{3}\cos t\space dt$
$\displaystyle=\int_0^{\frac{\pi}{2}} \sqrt{3}\sqrt{1-\sin^2 t}\cdot\sqrt{3}\cos t\space dt$
$\displaystyle=\int_0^{\frac{\pi}{2}} \sqrt{3}\cos t\cdot\sqrt{3}\cos t\space dt$
$\displaystyle=3\int_0^{\frac{\pi}{2}} \cos^2 t\space dt$
$\displaystyle=\frac{3}{2}\int_0^{\frac{\pi}{2}} 1+\cos 2t\space dt$
$\displaystyle=\frac{3}{2}\left[t+\frac{1}{2}\sin 2t\right]_0^{\frac{\pi}{2}}$
$\displaystyle=\frac{3}{4}\pi$(答え)