一个头疼的不定积分

$\int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}xdx$


不知道怎么算 分部后也不好算 凑微分貌似无力
已邀请:
$\displaystyle \int _0^1 \sqrt{\cfrac{1-x^2}{1+x^2}}\ x \ dx$
$\displaystyle =\cfrac{1}{2} \int _0^1 \sqrt{\cfrac{1-x^2}{1+x^2}} \ d(x^2)$ (设$x^2=u$)
$\displaystyle= \cfrac{1}{2}\int _0^1 \sqrt{\cfrac{1-u}{1+u}} \ du$
$\displaystyle= \cfrac{1}{2}\int _0^1 \cfrac{\sqrt{1-u^2}}{1+u} \ du$ (设$u=\textrm{sin}\theta$)
$\displaystyle= \cfrac{1}{2}\int _0^{\frac{\pi}{2}} \cfrac{\sqrt{1-\textrm{sin}^2\theta}}{1+\textrm{sin}\theta} \ d(\textrm{sin}\theta)$
$\displaystyle= \cfrac{1}{2}\int _0^{\frac{\pi}{2}} \cfrac{\textrm{cos}^2\theta}{1+\textrm{sin}\theta} \ d\theta$
$\displaystyle= \cfrac{1}{2}\int _0^{\frac{\pi}{2}} \cfrac{\left(1+\textrm{sin}\theta \right)\left(1-\textrm{sin}\theta \right)}{1+\textrm{sin}\theta} \ d\theta$
$\displaystyle= \cfrac{1}{2}\int _0^{\frac{\pi}{2}} 1-\textrm{sin}\theta \ d\theta$
$\displaystyle= \cfrac{1}{2}\left(\theta +\textrm{cos} \theta\right)|_0^\frac{\pi}{2}$
$=\cfrac{\pi}{4}-\cfrac{1}{2}.$

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