微积分(A)随缘一题[20]

试求：\(\int \sqrt{1-x^2}dx\)

\[\int \sqrt{1-x^2}dx \xlongequal[p^2+x^2=1]{p=\sqrt{1-x^2}}\int pdx \]

则 \(pdp+xdx=0\)，再考虑 \(I=\int pdx,J=\int xdp\)，则：

\[\begin{cases} I+J=\int pdx+xdp=\int d(px)=px+C=x\sqrt{1-x^2}+C \\ I-J=\int \frac{pdx-xdp}{p^2+x^2}=\int \frac{d\frac{x}{p}}{1+\left(\frac{x}{p}\right)^2}=\arctan \frac{x}{\sqrt{1-x^2}}+C=\arcsin x+C \end{cases} \]

所以：

\[\int \sqrt{1-x^2}dx=I=\frac{1}{2}\left( x\sqrt{1-x^2}+\arcsin x \right)+C \]