∫cos(2x)sec(3x)

方法1.\begin{align} \int\frac{\cos2x}{\cos3x}\mathrm d x &=\int\frac{\cos2x}{\cos3x\cos x}\mathrm d\sin x\\ &=\int\frac{\cos2x}{(-3+4\cos^2x)\cos^2x}\mathrm d\sin x\\ &=\int\frac{1-2t^2}{(1-4t^2)(1-t^2)}\mathrm d t\\ &=\frac16\int\left( \frac2{1-2t}+\frac2{1+2t}+\frac1{1+t}+\frac1{1-t} \right)\mathrm d t\\ &=\frac16\ln\left| \frac{(1+2t)(1+t)}{(1-2t)(1-t)} \right|+C\\ &=\frac16\ln\left| \frac{(1+2\sin x)(1+\sin x)}{(1-2\sin x)(1-\sin x)} \right|+C. \end{align}方法2. Mathematica

WolframAlpha["Integrate[Cos[2x]Sec[3x],x]", "PodCells", PodStates -> {"IndefiniteIntegral__Step-by-step solution"}][[2]]

$$\begin{array}{c}\text{Take the integral:}\\ \int \cos \left(2x\right)\sec \left(3x\right)\mathrm{d}x\\ \text{Write }\cos \left(2x\right)\sec \left(3x\right)\text{as }\frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}-\frac{{\sin }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\text{: }\\ \text{= }\int \left(\frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}-\frac{{\sin }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\right)\mathrm{d}x\\ \text{Integrate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants: }\\ \text{= }-\int \frac{{\sin }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{⁠}\text{⁠ }\text{Multiply }\text{numerator }\text{and }\text{denominator }\text{of }\frac{{\sin }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\text{by }-\cot \left(x\right){\csc }^3\left(x\right)\text{: }\\ \text{= }-\int -\frac{\cot \left(x\right)\csc \left(x\right)}{3{\cot }^2\left(x\right)-{\cot }^4\left(x\right)}\mathrm{d}x+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{⁠}\text{⁠ }\text{Prepare }\text{to }\text{substitute }u=\csc \left(x\right)\text{. }\text{Rewrite }-\frac{\cot \left(x\right)\csc \left(x\right)}{3{\cot }^2\left(x\right)-{\cot }^4\left(x\right)}\text{using }{\cot }^2\left(x\right)={\csc }^2\left(x\right)-1\text{: }\\ \text{= }-\int -\frac{\cot \left(x\right)\csc \left(x\right)}{-4+5{\csc }^2\left(x\right)-{\csc }^4\left(x\right)}\mathrm{d}x+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{For }\text{the }\text{integrand }-\frac{\cot \left(x\right)\csc \left(x\right)}{-4+5{\csc }^2\left(x\right)-{\csc }^4\left(x\right)}\text{, }\text{substitute }u=\csc \left(x\right)\text{and }\mathrm{d}u=-\cot \left(x\right)\csc \left(x\right)\mathrm{d}x\text{: }\\ \text{= }-\int \frac{1}{-u^4+5u^2-4}\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{For }\text{the }\text{integrand }\frac{1}{-u^4+5u^2-4}\text{, }\text{use }\text{partial }\text{fractions: }\\ \text{= }-\int \left(\frac{1}{6\left(u-1\right)}-\frac{1}{6\left(u+1\right)}+\frac{1}{12\left(u+2\right)}-\frac{1}{12\left(u-2\right)}\right)\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{Integrate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants: }\\ \text{= }-\frac{1}{12}\int \frac{1}{u+2}\mathrm{d}u+\frac{1}{6}\int \frac{1}{u+1}\mathrm{d}u-\frac{1}{6}\int \frac{1}{u-1}\mathrm{d}u+\frac{1}{12}\int \frac{1}{u-2}\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{For }\text{the }\text{integrand }\frac{1}{u+2}\text{, }\text{substitute }s=u+2\text{and }\mathrm{d}s=\mathrm{d}u\text{: }\\ \text{= }-\frac{1}{12}\int \frac{1}{s}\mathrm{d}s+\frac{1}{6}\int \frac{1}{u+1}\mathrm{d}u-\frac{1}{6}\int \frac{1}{u-1}\mathrm{d}u+\frac{1}{12}\int \frac{1}{u-2}\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{The }\text{integral }\text{of }\frac{1}{s}\text{is }\log \left(s\right)\text{: }\\ \text{= }-\frac{\log \left(s\right)}{12}+\frac{1}{6}\int \frac{1}{u+1}\mathrm{d}u-\frac{1}{6}\int \frac{1}{u-1}\mathrm{d}u+\frac{1}{12}\int \frac{1}{u-2}\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{For }\text{the }\text{integrand }\frac{1}{u+1}\text{, }\text{substitute }p=u+1\text{and }\mathrm{d}p=\mathrm{d}u\text{: }\\ \text{= }-\frac{\log \left(s\right)}{12}+\frac{1}{6}\int \frac{1}{p}\mathrm{d}p-\frac{1}{6}\int \frac{1}{u-1}\mathrm{d}u+\frac{1}{12}\int \frac{1}{u-2}\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{The }\text{integral }\text{of }\frac{1}{p}\text{is }\log \left(p\right)\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}-\frac{1}{6}\int \frac{1}{u-1}\mathrm{d}u+\frac{1}{12}\int \frac{1}{u-2}\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{For }\text{the }\text{integrand }\frac{1}{u-1}\text{, }\text{substitute }w=u-1\text{and }\mathrm{d}w=\mathrm{d}u\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}-\frac{1}{6}\int \frac{1}{w}\mathrm{d}w+\frac{1}{12}\int \frac{1}{u-2}\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{The }\text{integral }\text{of }\frac{1}{w}\text{is }\log \left(w\right)\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}-\frac{\log \left(w\right)}{6}+\frac{1}{12}\int \frac{1}{u-2}\mathrm{d}u+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{For }\text{the }\text{integrand }\frac{1}{u-2}\text{, }\text{substitute }v=u-2\text{and }\mathrm{d}v=\mathrm{d}u\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}-\frac{\log \left(w\right)}{6}+\frac{1}{12}\int \frac{1}{v}\mathrm{d}v+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{The }\text{integral }\text{of }\frac{1}{v}\text{is }\log \left(v\right)\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\int \frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\mathrm{d}x\\ \text{⁠}\text{⁠ }\text{Multiply }\text{numerator }\text{and }\text{denominator }\text{of }\frac{{\cos }^2\left(x\right)}{{\cos }^3\left(x\right)-3{\sin }^2\left(x\right)\cos \left(x\right)}\text{by }-{\csc }^2\left(x\right)\sec \left(x\right)\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\int -\frac{\cot \left(x\right)\csc \left(x\right)}{3-{\cot }^2\left(x\right)}\mathrm{d}x\\ \text{⁠}\text{⁠ }\text{Prepare }\text{to }\text{substitute }{z}_1=\csc \left(x\right)\text{. }\text{Rewrite }-\frac{\cot \left(x\right)\csc \left(x\right)}{3-{\cot }^2\left(x\right)}\text{using }{\cot }^2\left(x\right)={\csc }^2\left(x\right)-1\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\int -\frac{\cot \left(x\right)\csc \left(x\right)}{4-{\csc }^2\left(x\right)}\mathrm{d}x\\ \text{For }\text{the }\text{integrand }-\frac{\cot \left(x\right)\csc \left(x\right)}{4-{\csc }^2\left(x\right)}\text{, }\text{substitute }{z}_1=\csc \left(x\right)\text{and }\mathrm{d}{z}_1=-\cot \left(x\right)\csc \left(x\right)\mathrm{d}x\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\int \frac{1}{4-z_1^2}\mathrm{d}{z}_1\\ \text{Factor }4\text{from }\text{the }\text{denominator: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\int \frac{1}{4\left(1-\frac{z_1^2}{4}\right)}\mathrm{d}{z}_1\\ \text{Factor }\text{out }\text{constants: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\frac{1}{4}\int \frac{1}{1-\frac{z_1^2}{4}}\mathrm{d}{z}_1\\ \text{For }\text{the }\text{integrand }\frac{1}{1-\frac{z_1^2}{4}}\text{, }\text{substitute }{z}_2=\frac{z_1}{2}\text{and }\mathrm{d}{z}_2=\frac{1}{2}\mathrm{d}{z}_1\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\frac{1}{2}\int \frac{1}{1-z_2^2}\mathrm{d}{z}_2\\ \text{The }\text{integral }\text{of }\frac{1}{1-z_2^2}\text{is }{\tanh }^{-1}\left(z_2\right)\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\frac{1}{2}{\tanh }^{-1}\left(z_2\right)+\text{constant }\\ \text{Substitute }\text{back }\text{for }{z}_2=\frac{z_1}{2}\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\frac{1}{2}{\tanh }^{-1}\left(\frac{z_1}{2}\right)+\text{constant }\\ \text{Substitute }\text{back }\text{for }{z}_1=\csc \left(x\right)\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{\log \left(v\right)}{12}-\frac{\log \left(w\right)}{6}+\frac{1}{2}{\coth }^{-1}\left(2\sin \left(x\right)\right)+\text{constant }\\ \text{Substitute }\text{back }\text{for }v=u-2\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{1}{12}\log \left(u-2\right)-\frac{\log \left(w\right)}{6}+\frac{1}{2}{\coth }^{-1}\left(2\sin \left(x\right)\right)+\text{constant }\\ \text{Substitute }\text{back }\text{for }w=u-1\text{: }\\ \text{= }\frac{\log \left(p\right)}{6}-\frac{\log \left(s\right)}{12}+\frac{1}{12}\log \left(u-2\right)-\frac{1}{6}\log \left(u-1\right)+\frac{1}{2}{\coth }^{-1}\left(2\sin \left(x\right)\right)+\text{constant }\\ \text{Substitute }\text{back }\text{for }p=u+1\text{: }\\ \text{= }-\frac{\log \left(s\right)}{12}+\frac{1}{12}\log \left(u-2\right)-\frac{1}{6}\log \left(u-1\right)+\frac{1}{6}\log \left(u+1\right)+\frac{1}{2}{\coth }^{-1}\left(2\sin \left(x\right)\right)+\text{constant }\\ \text{Substitute }\text{back }\text{for }s=u+2\text{: }\\ \text{= }\frac{1}{12}\log \left(u-2\right)-\frac{1}{6}\log \left(u-1\right)+\frac{1}{6}\log \left(u+1\right)-\frac{1}{12}\log \left(u+2\right)+\frac{1}{2}{\coth }^{-1}\left(2\sin \left(x\right)\right)+\text{constant }\\ \text{Substitute }\text{back }\text{for }u=\csc \left(x\right)\text{: }\\ \text{= }\frac{1}{12}\log \left(\csc \left(x\right)-2\right)-\frac{1}{6}\log \left(\csc \left(x\right)-1\right)+\frac{1}{6}\log \left(\csc \left(x\right)+1\right)-\frac{1}{12}\log \left(\csc \left(x\right)+2\right)+\frac{1}{2}{\coth }^{-1}\left(2\sin \left(x\right)\right)+\text{constant }\\ \text{Factor }\text{the }\text{answer }\text{a }\text{different }\text{way: }\\ \text{= }\frac{1}{12}\left(\log \left(\csc \left(x\right)-2\right)-2\log \left(\csc \left(x\right)-1\right)+2\log \left(\csc \left(x\right)+1\right)-\log \left(\csc \left(x\right)+2\right)+6{\coth }^{-1}\left(2\sin \left(x\right)\right)\right)+\text{constant }\\ \text{An }\text{alternative }\text{form }\text{of }\text{the }\text{integral }\text{is: }\\ \text{= }\frac{1}{12}\left(\log \left(\frac{\csc \left(x\right)-2}{\csc \left(x\right)+2}\right)+6{\coth }^{-1}\left(2\sin \left(x\right)\right)+4{\coth }^{-1}\left(\csc \left(x\right)\right)\right)+\text{constant }\end{array}$$ 方法3. 使用欧拉公式,$\cos2x\sec3x=\frac{e^{2ix}+e^{-2ix}}{e^{3ix}+e^{-3ix}}$.作代换$u=e^{-ix},\mathrm dx=iu\mathrm du$.
$\int\frac{u^2+u^{-2}}{u^3+u^{-3}}u^{-1}~\mathrm du=\int\frac{u^4+1}{u^6+1}~\mathrm du=\frac23\int\frac{\mathrm du}{u^2+1}+\frac13\int\frac{u^2+1}{u^4-u^2+1}\mathrm du=\frac23\arctan u+\frac13\arctan\frac u{1-u^2}+C$.
所以$\int\cos2x\sec3x~\mathrm dx=\frac{2i}3\arctan e^{-ix}+\frac i3\arctan\frac{e^{-ix}}{1-e^{-2ix}}+C$.
如果利用恒等式$i\arctan x=\frac12\log\frac{1+x}{1-x}$化简的话,应该会得出,和上面的结果是相等的.