用来保存考研常用的积分公式，必须要牢牢记住
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三角函数相关
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\(\int tanxdx = \int \frac{sinxdx}{cosx} = \int \frac{-d(cosx)}{cosx}=-\ln \left | cosx \right | +C\)
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\(\int cotxdx = \int \frac{cosxdx}{sinx} = \int \frac{d(sinx)}{sinx}=\ln \left | sinx \right | +C\)
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\(\int \sec xdx=\int \frac{1}{\cos x}dx=\int \frac{ \frac{sinx+1}{cos^{2}x} }{\frac{sinx+1}{cosx}}dx = \int \frac{\frac{sinx}{cos^{2}x} + \frac{1}{cos^{2}x} }{secx+tanx}dx = \int \frac{d(secx+tanx)}{secx+tanx} = ln|secx+tanx| + C\)
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\(\int \csc xdx=\int \frac{1}{\sin x}dx=\int \frac{ \frac{1-cosx}{sin^{2}x} }{\frac{1-cosx}{sinx}}dx = \int \frac{\frac{-cosx}{sin^{2}x} + \frac{1}{sin^{2}x} }{cscx-cotx}dx = \int \frac{d(cscx - cotx)}{cscx-cotx} = ln|cscx-cotx| + C\)
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\(\int \frac{dx}{sin^{2}x}=\int -d(\frac{cosx}{sinx})=-cotx+C\)
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\(\int \frac{dx}{cos^{2}x}=\int d(\frac{sinx}{cosx})=tanx+C\)
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\(\int secx\cdot tanxdx = \int \frac{sinx}{cos^{2}x}dx =\int \frac{-d(cosx))}{cos^{2}x}=\frac{1}{cosx}+C=secx+C\)
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\(\int cscx\cdot cotxdx = \int \frac{cosx}{sin^{2}x}dx =\int \frac{d(sinx))}{sin^{2}x}=\frac{-1}{sinx}+C=-cscx+C\)
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当n为正奇数的时候
\(\int_{0}^{\frac{\pi}{2}}sin^{n}xdx=\int_{0}^{\frac{\pi}{2}}cos^{n}xdx=\frac{n-1}{n}\cdot\frac{n-3}{n-2} \cdot \cdot \cdot \cdot \frac{2}{3}\cdot 1\)
当n为正偶数的时候：
\(\int_{0}^{\frac{\pi}{2}}sin^{n}xdx=\int_{0}^{\frac{\pi}{2}}cos^{n}xdx=\frac{n-1}{n}\cdot\frac{n-3}{n-2} \cdot \cdot \cdot \cdot \frac{1}{2}\cdot \frac{\pi}{2}\)
多项式
- x 与 a的平方和 \(x=atan\theta\)
\(\int \frac{dx}{a^{2}+x^{2}}=\int \frac{d(atan\theta))}{a^{2}+a^{2}tan^{2}\theta}=\int \frac{d\theta}{a}=\theta+C=arctan\frac{x}{a}+C\)
\(\int \frac{dx}{\sqrt{a^{2}+x^{2}}} = \int \frac{d(atan\theta ) }{\sqrt{a^{2}+a^{2}tan^{2}\theta}}=\int \frac{d\theta}{cos\theta}=ln|cos\theta +tan\theta|+C=ln(\frac{\sqrt{a^{2}+x^{2}}}{a}+\frac{x}{a})+C=ln(\sqrt{a^{2}+x^{2}}+x)+{C}‘\)
\(\int \frac{dx}{\sqrt{x^{2}-a^{2}}}=ln(\sqrt{x^{2}-a^{2}}+x)+C\)
- x平方 - a的平方 \(x = \frac{a}{sin\theta}\)
\(\sqrt{x^{2} - a^{2}} = \sqrt{\frac{a^2(1-sin^{2}\theta))}{sin^{2}\theta}} = \sqrt{a^{2}\frac{cos^{2}\theta}{sin^{2}\theta}}=acot\theta\)
\(\int \frac{dx}{x^{2}-a^{2}} = \frac{1}{2a}\int \frac{(x+a)-(x-a)}{(x+a)(x-a)}dx=\frac{1}{2a}\int (\frac{1}{x-a} - \frac{1}{x+a})=ln|\frac{x-a}{x+a}|+C\)
\(\int \frac{dx}{a^{2}-x^{2}} = \frac{1}{2a}\int \frac{(a+x)+(a-x)}{(a+x)(a-x)}dx=\frac{1}{2a}\int (\frac{1}{a-x} - \frac{1}{a+x})=ln\frac{a-x}{a+x}+C\)
\(\int \sqrt{x^{2} - a^{2}} dx =\frac{x}{2}\sqrt{x^{2} - a^{2}}-\frac{a^{2}}{2}ln|x+\sqrt{x^{2} - a^{2}}|+C\)
\(\int \sqrt{x^{2} + a^{2}} dx =\frac{x}{2}\sqrt{x^{2} + a^{2}}+\frac{a^{2}}{2}ln|x+\sqrt{x^{2} + a^{2}}|+C\)
\(\int \frac{dx}{\sqrt{a^{2} - x^{2}} } =arcsin\frac{x}{a}+C\)
\(\int \sqrt{a^{2} - x^{2}}dx =\frac{x}{2}\sqrt{a^{2} - x^{2}}+\frac{a^{2}}{2}arcsin\frac{x}{a}+C\)