微积分小题集(2)
微积分小题集(2)
\(\newcommand \d{\ \mathrm{d}} \newcommand \f{\int}\newcommand \dx{\ \mathrm{d}x}\)
证明 \(\lim_{x \to \infty}(\sin \sqrt{x^2+2}-\sin \sqrt {x^2+1})=0\)
\(\sin x - \sin y \le |x - y|\) 即可得证。
\[\int \frac{x^2}{1+x^2} \d x = \int (1-\frac 1{1+x^2})\d x=x - \arctan x+ C
\]
\[\int \frac{x^2}{1-x^2} \d x = \int (-1+\frac 1{1-x^2})\d x=-x+\frac 12\ln |\frac{1+x}{1-x}|+C
\]
\[\int (2^x + 3 ^ x)^2 \d x = \int (4^x + 9^x+2\cdot 6^x) \d x=\frac{4^x}{\ln 4}+\frac{9^x}{\ln 9}+2\frac{6^x}{\ln 6}+C
\]
\[\int \sqrt{1 - \sin 2x} \d x = \int \sqrt{(\cos x - \sin x)^2}=sgn(\cos x - \sin x)(\sin x + \cos x) + C
\]
\[\int \tan^2x \d x=\int (\tan ^2 x + 1) - 1 \d x=\tan x - x +C
\]
\[\f \frac{\dx}{1 - x^2} = \frac12 \f \frac{1}{1+x}+\frac{1}{1-x}\dx = \frac 12 \ln|\frac{1+x}{1-x}|+C
\]
推论：\(\f \frac{\dx}{a^2 - x^2}=\frac{1}{2a}\ln|\frac{a+x}{a-x}|+C\)
\[\f \frac{\dx }{ \sqrt{x^2 \pm 1}} = \ln|x + \sqrt{x^2 \pm 1}|+C
\]
\[\f \frac{\d x}{\sin^2 x} = -\cot x + C
\]
\[\f \frac{\dx}{x \sqrt{x^2 + 1}}=\f \frac{\dx}{x |x|\sqrt{1 + \frac{1}{x^2}}}=-\f \frac{\d\frac 1{|x|}}{\sqrt{1 + \frac{1}{x^2}}}=-\ln\left|\frac 1{|x|}+\sqrt{1+\frac 1{x^2}}\right|+C
\]
\[\f \frac{\dx}{\sqrt{x(1-x)}}=2\f \frac{\d \sqrt{x}}{\sqrt {1-x}}=2 \arcsin\sqrt x+C
\]
\[\f \frac{\dx}{\sqrt{1 + e^{2x}}}=-\f \frac{\d e^{-x}}{\sqrt{1 + e^{-2x}}} = - \ln|e^{-x}+\sqrt{1 + e^{-2x}}| +C
\]
\[\f \frac{\dx}{x \ln x \ln (\ln x)}=\f \frac{\d \ln x}{\ln x \ln (\ln x)}=\f \frac{\d \ln (\ln x)}{\ln (\ln x)} = \ln |\ln \ln x|+C
\]
\[\f \frac{x^{14} \d x}{(x^5 + 1)^4} = \f \frac{x^{14} \d x}{x^{20}(x^{-5} + 1)^4}=-\frac 1{5}\f \frac{\d (x^{-5}+1)}{(x^{-5} + 1)^4}=-\frac 1{15}\frac{1}{(x^{-5}+1)^3}+C
\]
\[\f x^3 \sqrt[3]{1 + x^2} \dx =\frac 12\f (1 + x^2 - 1)(1+x^2)^{\frac13} \d (1 + x^2) = \frac {3(1+x^2)^{\frac 73}}{14}-\frac{3(1 + x^2)^{\frac 43}}{8}+C
\]
令 \(u = \sin x\)
\[\begin{align}
&\f \sqrt{1-x^2} \d x\\
=&\f \cos ^2u \d u\\
=&\f \frac{\cos 2u+1}{2} \d u\\
=& (\frac{\sin 2u}{4}+\frac{u}{2})\\
=& \frac12(x\sqrt {1-x^2}+\arcsin x) +C
\end{align}
\]
令 \(x = \tan t,u = \sin t\)
\[\begin{align}
&\f \sqrt{1 + x^2} \dx \\
=& \f \frac{1}{\cos^3 t}\d t \\
=& \f \frac{1}{(1-\sin^2 t)^2} \d \sin t\\
=& \f \frac 14(\frac{1}{1 + u}+\frac{1}{1-u})^2 \d u\\
=& -\frac 14(\frac{1}{1 + u}-\frac{1}{1-u}) +\frac 12\f \frac{1}{(1+u)(1-u)}\d u\\
=& -\frac 14(\frac{1}{1 + u}-\frac{1}{1-u}) + \frac 14 \ln|\frac{1+u}{1-u}|+C\\
=&\frac12 x\sqrt{1 + x^2} + \frac 14 \ln|1+2x^2 + 2x \sqrt{1 + x^2}|+C\\
=&\frac12 x\sqrt{1 + x^2} + \frac 12 \ln|x+\sqrt{1 + x^2}|+C\\
\end{align}
\]
其中将 \(u = \sin t = \sqrt{\frac{x^2}{1 + x^2}}\)，带入。
\[\begin{align}
&\f \sqrt{x^2-1} \dx = x \sqrt{x^2 - 1} - \f \frac{(x^2-1)+1}{\sqrt{x^2-1}} \d x\\
&\f \sqrt{x^2-1} \dx = \frac{x\sqrt{x^2 - 1} - \f \frac{1}{\sqrt{x^2-1}} \d x}{2}=\frac 12(x\sqrt{x^2-1}-\ln |x + \sqrt{x^2 - 1}|)+C\\
\end{align}
\]
\[\begin{align}
F(n)&=\f \frac{\dx}{(1+x^a)^{n}}\dx \\
&=\frac{x}{(1+x^a)^{n}}+an\f\frac{x^{a}\dx}{(1+x^a)^{n+1}}\\
&=\frac{x}{(1+x^a)^{n}}+an\f\frac{(1+x^{a})-1\dx}{(1+x^a)^{n+1}}\\
&=\frac{x}{(1+x^a)^{n}}+an(F(n)-F(n+1))\\
F(n+1) &= \frac{(an-1)F(n) + \frac{x}{(1+x^a)^n}}{an}
\end{align}
\]
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