积分方法收集

积分方法收集

\[\Gamma(n)=\int^{+\infty}_0x^{n-1}e^{-x}dx\]
\[\Gamma(n)=(n-1)!\]
\[\Gamma(\frac{1}{2})=\sqrt{\pi}\]

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\[\int_0^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi} }{2}\]

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\[\int xe^xdx=\int (xe^x+e^x-e^x)dx=(x-1)e^x+C\]

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\[\int \frac{dx}{\sin x}=\int \frac{dx}{2\sin \frac{x}{2}\cos \frac{x}{2} }=\int \frac{\sec^2 \frac{x}{2} d \frac{x}{2} }{\tan \frac{x}{2} }=\int \frac{d\tan \frac{x}{2} }{\tan \frac{x}{2} }=\ln (\tan \frac{x}{2})+C\]

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\[\int \frac{1}{1+e^x}dx=\int \frac{e^{-x} }{1+e^{-x} }dx=-\int \frac{1}{1+e^{-x} }d(1+e^{-x})=-\ln (1+e^{-x})+C\]

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\[\int \sqrt{\frac{1-t}{1+t} }dt\to令\frac{1-t}{1+t}=u^2\to t^2=\frac{1-u}{1+u}\]
\[dt=\frac{-4u}{(1+u^2)^2}du\to \int \frac{-4u^2}{(1+u^2)^2}du=-4\int(\frac{1}{1+u^2}-\frac{1}{(1+u^2)^2})du\]
后面部分令\(u=\tan\theta\)，即可计算出。

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