不定积分例题训练
precedent 1
\[\begin{eqnarray}
\int\frac{dx}{\sin^{2}x\cos^{2}x}=?
\\ \\
\Rightarrow\int\frac{1}{\sin^{2}x\cos^{2}x}dx
\\ \\
\because\sin^{2}\beta+\cos^{2}\beta=1
\\ \\
\therefore\int\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}dx
\\ \\
\int\left(\frac{\sin^{2}x}{\sin^{2}x\cos^{2}x}+\frac{\cos^{2}x}{\sin^{2}x\cos^{2}x}\right)dx
\\ \\
\int\left(\frac{1}{\cos^{2}x}+\frac{1}{\sin^{2}x}\right)dx
\\ \\
\because\frac{1}{\cos x}=\sec x,\enspace \sec^{2}x=(\tan x)^{\prime}
\\ \\
\because \int \frac{1}{\sin^{2} x}dx=-\cot x+C
\\ \\
结果: \enspace \tan x-\cot x+C
\end{eqnarray}
\]
precedent 0
\[\begin{eqnarray}
\int\sin^{2}\frac{x}{2}dx=?
\\ \\
前知识: \enspace \sin^{2}\beta=\frac{1-\cos2\beta}{2}
\\ \\
\therefore\int\left[\frac{1-\cos\left(\frac{x}{2}\cdot2\right)}{2}\right]dx
\\ \\
\int\left(\frac{1}{2}-\frac{\cos x}{2}\right)dx
\\ \\
\int\frac{1}{2}\left(1-\cos x\right)dx
\\ \\
\frac{1}{2}\int1dx-\frac{1}{2}\int\cos xdx
\\ \\
结果: \enspace \frac{1}{2}x-\frac{1}{2}\sin x+C
\end{eqnarray}
\]
exercise 1
\[\begin{eqnarray}
\int \frac{1}{\sqrt[]{x}}dx
\newline \newline
\int \frac{1}{\sqrt[]{x}}dx=\int \frac{1}{x^{\frac{1}{2}}}dx=\int x^{-\frac{1}{2}}dx
\newline \newline
\Rightarrow \frac{1}{-\frac{1}{2}+1}x^{-\frac{1}{2}+1}+C
\newline \newline
2 \times x^{\frac{1}{2}}+C
\newline \newline
2\sqrt[]{x} +C
\end{eqnarray}
\]
exercise 2
\[
\begin{align}
\int x^{2} \times \sqrt[3]{x} dx
\\ \\
\because x^{2} \times \sqrt[3]{x} \Rightarrow x^{2+\frac{1}{3} }=x^{\frac{7}{3} }
\\ \\
\therefore \int x^{\frac{7}{3} }dx= \frac{1}{\frac{7}{3}+1} x^{\frac{7}{3} +1 } + C
\\ \\
\Rightarrow \frac{3}{10} x^{\frac{10}{3} } + C
\end{align}
\]
exercise 3
\[\begin{align}
\int(\sqrt{x}+1)(x-\sqrt{x}) d x=?
\\ \\
\Rightarrow \int\left(x^{\frac{1}{2}}+1\right)\left(x-x^{-\frac{1}{2}}\right) d x
\\ \\
= \int(x^{\frac{3}{2}}-1+x-x^{-\frac{1}{2}} )dx
\\ \\
\int x^{\frac{3}{2}} d x=\frac{1}{\frac{3}{2}+1} x^{\frac{3}{2}+1}=\frac{2}{5} x^{\frac{5}{2}}
\\ \\
\int x d x=\frac{1}{1+1} x^{1+1}=\frac{1}{2} x^{2}
\\ \\
\int x^{-\frac{1}{2}} d x=\frac{1}{-\frac{1}{2}+1} x^{-\frac{1}{2}+1}=2 x^{\frac{1}{2}}
\\ \\
\int 1 d x=x
\\ \\
\Rightarrow \frac{2}{5} x^{\frac{5}{2}}-x+\frac{1}{2} x^{2}-2 x^{\frac{1}{2}}+C
\end{align}
\]
exercise 4
\[
\begin{align}
\int\frac{2^{x+1}-5^{x-1}}{10^{x}}dx=?
\\ \\
\Rightarrow\int\left(\frac{2^{x+1}}{10^{x}}-\frac{5^{x-1}}{10^{x}}\right)dx
\\ \\
1标: \int \left(\frac{2^{x}\cdot2}{10^{x}}-\frac{5^{x}\cdot5^{-1}}{10^{x}}\right)dx
\\ \\
2标: \int\left(2\frac{2^{x}}{10^{x}}-\frac{1}{5}\frac{5^{x}}{10^{x}}\right)dx
\\ \\
3标: \int\left[2\left(\frac{1}{5}\right)^{x}-\frac{1}{5}\left(\frac{1}{2}\right)^{x}\right]dx
\\ \\
\Rightarrow2\int\left(\frac{1}{5}\right)^{x}dx-\frac{1}{5}\int\left(\frac{1}{2}\right)^{x}dx
\\ \\
\because \int a^{x}dx=\frac{a^{x}}{\ln a}+C
\\ \\
得: 2\frac{(\frac{1}{5})^{x}}{\ln(\frac{1}{5})}-
\frac{1}{5}\frac{\left(\frac{1}{2}\right)^{x}}{\ln(\frac{1}{2})}+C
\end{align}
\]
exercise 5
\[\begin{align}
\int \frac{x^{2}}{1+x^{2}}dx=?
\\ \\
\Rightarrow\int\frac{x^{2}+1-1}{1+x^{2}}dx
\\ \\
\int(\frac{x^{2}+1}{1+x^{2}}-\frac{1}{1+x^{2}})dx
\\ \\
\int(1-\frac{1}{1+x^{2}})dx
\\ \\
\because (x)^{\prime}=1,
\enspace
(\arctan x)^{\prime}=\frac{1}{1+x^{2}}
\\ \\
\therefore\int\frac{x^{2}}{1+x^{2}}dx=x-\arctan x+C
\\ \\
\end{align}
\]
exercise 6
\[\begin{align}
\int\frac{x}{\sqrt{1+x}}dx=?
\\ \\
设:\sqrt{1+x}=u,\enspace 则: x=u^{2}-1
\\ \\
\Rightarrow\int\frac{u^{2}-1}{u}d(u^{2}-1)
\\ \\
d(u^{2}-1)=du\cdot(u^{2}-1)^{\prime}=du\cdot2u
\\ \\
\Rightarrow\int\frac{u^{2}-1}{u}\cdot2udu=2\int\frac{u^{2}-1}{u}\cdot udu
\\ \\
=2\int(u^{2}-1)du=2\int u^{2}du-2\int du
\\ \\
2\int u^{2}du=2(\frac{1}{2+1}u^{2+1})+C=\frac{2}{3}u^{3}+C
\\ \\
2\int du=2u+C
\\ \\
\boxed{ \frac{2}{3}u^{3}-2u+C }
\\ \\
\frac{2}{3}(\sqrt{1+x})^{3}-2\sqrt{1+x}=2[\frac{1}{3}(1+x)\sqrt{1+x}-\sqrt{1+x}]
\\ \\
=2\sqrt{1+x}[\frac{1}{3}(1+x)-1]+C
\end{align}
\]
