问题及解答

计算定积分 $\displaystyle\int_0^1\sqrt{\frac{1-t}{1+t}}\mathrm{d}t$.

\[
\begin{split}
\int_0^1\sqrt{\frac{1-t}{1+t}}\mathrm{d}t&=\int_0^1\frac{1-t}{\sqrt{1-t^2}}\mathrm{d}t\\
&=\int_0^1\frac{1}{\sqrt{1-t^2}}\mathrm{d}t-\int_0^1\frac{t}{\sqrt{1-t^2}}\mathrm{d}t\\
&=\arcsin t\Bigr|_0^1 - \int_0^1\frac{\frac{1}{2}\mathrm{d}t^2}{\sqrt{1-t^2}}=\frac{\pi}{2}-\frac{1}{2}\int_0^1\frac{\mathrm{d}u}{\sqrt{1-u}}\\
&=\frac{\pi}{2}-\frac{1}{2}\cdot((-2)\sqrt{1-u})\Bigr|_0^1=\frac{\pi}{2}+\sqrt{1-u}\Bigr|_0^1\\
&=\frac{\pi}{2}-1\ .
\end{split}
\]

若令 $x=\sqrt{\frac{1-t}{1+t}}$, 则 $x^2=\frac{1-t}{1+t}$, 这推出 $t=\frac{1-x^2}{1+x^2}$. 从而

\[
\mathrm{d}t=\mathrm{d}\frac{1-x^2}{1+x^2}=\frac{-2x\mathrm{d}x\cdot(1+x^2)-(1-x^2)\cdot 2x\mathrm{d}x}{(1+x^2)^2}=\frac{-4x}{(1+x^2)^2}\mathrm{d}x.
\]

因此,

\[
\int_0^1\sqrt{\frac{1-t}{1+t}}\mathrm{d}t=\int_1^0 x\cdot\frac{-4x}{(1+x^2)^2}\mathrm{d}x=4\int_0^1\frac{x^2}{(1+x^2)^2}\mathrm{d}x=4\int_0^1\frac{1}{1+x^2}\mathrm{d}x-4\int_0^1\frac{1}{(1+x^2)^2}\mathrm{d}x\tag{*}
\]

而计算积分 $\displaystyle\int_0^1\frac{1}{(1+x^2)^2}\mathrm{d}x$ 我们通常需要用分部积分法. 一般的, 对于 $I_n=\displaystyle\int_0^1\frac{1}{(1+x^2)^n}\mathrm{d}x$ 通过分部积分可导出递推公式.

\[
\begin{split}
\int\frac{1}{1+x^2}\mathrm{d}x&=\frac{x}{1+x^2}\biggr|_0^1-\int_0^1 x\mathrm{d}\frac{1}{1+x^2}\\
&=\frac{1}{2}-\int_0^1 x\cdot\frac{-1}{(1+x^2)^2}\cdot 2x\mathrm{d}x\\
&=\frac{1}{2}+2\int_0^1\frac{x^2}{(1+x^2)^2}\mathrm{d}x\\
&=\frac{1}{2}+2\int_0^1\Bigl[\frac{1}{1+x^2}-\frac{1}{(1+x^2)^2}\Bigr]\mathrm{d}x\\
&=\frac{1}{2}+2\int_0^1\frac{1}{1+x^2}\mathrm{d}x-2\int_0^1\frac{1}{(1+x^2)^2}\mathrm{d}x\\
\end{split}
\]

这推出

\[
2\int_0^1\frac{1}{(1+x^2)^2}\mathrm{d}x=\int_0^1\frac{1}{1+x^2}\mathrm{d}x+\frac{1}{2}=\arctan x\biggr|_0^1+\frac{1}{2}=\frac{\pi}{4}+\frac{1}{2},
\]

因此

\[
\int_0^1\frac{1}{(1+x^2)^2}\mathrm{d}x=\frac{\pi}{8}+\frac{1}{4}.
\]

代入 (*), 得

\[
\int_0^1 \sqrt{\frac{1-t}{1+t}}\mathrm{d}t=4\cdot\arctan x\biggr|_0^1-4\cdot\Bigl(\frac{\pi}{8}+\frac{1}{4}\Bigr)=\frac{\pi}{2}-1.
\]