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幾道積分練習題

 

1. \displaystyle\int\frac{\sin(x)\cos(x)}{\sin(x)+\cos(x)}\dx

 

 解1

 

    \begin{align*} &\,\int\frac{\sin(x)\cos(x)}{\sin(x)+\cos(x)}\dx\\[1mm] =&\,\frac{1}{\sqrt{2}}\int\frac{\sin(x)\cos(x)}{\sin(x+\frac{\pi}{4})}\dx\\[1mm] =&\,\frac{1}{2\sqrt{2}}\frac{\Big[\sin(u)-\cos(u)\Big]\Big[\cos(u)+\sin(u)\Big]}{\sin(u)}\du\\[1mm] =&\,\frac{1}{2\sqrt{2}}\int\frac{2\sin^2(u)-1}{\sin(u)}\du\\[1mm] =&\,\frac{1}{\sqrt{2}}\int\sin(u)\du -\frac{1}{2\sqrt{2}}\int\csc(u)\du\\[1mm] =&\,-\frac{\cos(x+\frac{\pi}{4})}{\sqrt{2}}\\[1mm] &\,\;-\frac{1}{2\sqrt{2}}\ln\abs{\csc(x+\frac{\pi}{4}) -\cot(x+\frac{\pi}{4})}+C \end{align*}

 

 解2

 

    \begin{align*} &\,\int\frac{\sin(x)\cos(x)}{\sin(x)+\cos(x)}\dx\\[1mm] =&\,\frac{1}{2}\int\frac{\big(\sin(x)+\cos(x)\big)^2-1}{\sin(x)+\cos(x)}\dx\\[1mm] =&\,\frac{1}{2}\int\big(\sin(x)+\cos(x)\big)\dx\\[1mm] &\,\quad\frac{1}{2\sqrt{2}}\int\frac{1}{\sin(x+\frac{\pi}{4})}\dx\\[1mm] =&\,\frac{1}{2}\big(-\cos(x)+\sin(x)\big)\\[1mm] &\,\quad-\frac{1}{2\sqrt{2}}\ln\abs{\csc(x+\frac{\pi}{4}) -\cot(x+\frac{\pi}{4})}+C \end{align*}

 

 

2. \displaystyle\int e^{\sin(x)}\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}\dx

 

 解1

 

由於被積分函數是指數函數乘上一坨東西,所以猜測答案的形式亦是

(1)   \begin{align*} e^{\sin(x)}g(x) \end{align*}

於是

    \begin{align*} \Big(e^{\sin(x)}g(x)\Big)' =&\,e^{\sin(x)}\cos(x)g(x)+e^{\sin(x)}g'(x)\\[1mm] =&\,e^{\sin(x)}\Big[\cos(x)g(x)+g'(x)\Big] \end{align*}

接下來解

    \begin{align*} \cos(x)g(x)+g'(x)=&\,\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}\\[1mm] =&\,x\cos(x)-\tan(x)\sec(x) \end{align*}

g(x)=x是明顯說不通的,所以無中生有一下

    \begin{align*} \cos(x)g(x)+g'(x)=&\,x\cos(x)-1+1-\tan(x)\sec(x)\\[1mm] =&\,\cos(x)\Big[x-\sec(x)\Big]+\Big[1-\tan(x)\sec(x)\Big] \end{align*}

成功地求出 g(x)=x-\sec(x) ,故答案為 e^{\sin(x)}\Big(x-\sec(x)\Big)+C

 

 解2

 

    \begin{align*} &\,\int e^{\sin(x)}\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}\dx\\[1mm] =&\,\int xe^{\sin(x)}\cos(x)\dx\\[1mm] &\,\quad-\int e^{\sin(x)} \tan(x)\sec(x)\dx\\[1mm] =&\,xe^{\sin(x)}-\int e^{\sin(x)}\dx\\[1mm] &\,\quad-e^{\sin(x)}\sec(x) +\int e^{\sin(x)}\dx\\[1mm] =&\,e^{\sin(x)}\big(x-\sec(x)\big)+C \end{align*}

 

 

3. \displaystyle\int\frac{\ln(x)}{(1+x^2)^{\frac{3}{2}}}\dx

 

 解1

 

x=\sinh(t) , \dx=\cosh(t)\dt ,則

    \begin{align*} &\,\int \frac{\ln\big(\sinh(t)\big)}{\cosh^3(t)}\cosh(t)\dt\\[1mm] =&\,\int\mathrm{sech}^2(t)\ln\big(\sinh(t)\big)\dt\\[1mm] =&\,\tanh(t)\ln\big(\sinh(t)\big) -\int\underbrace{\tanh(t)\cdot\frac{\cosh(t)}{\sinh(t)}}_{=1}\dt\\[1mm] =&\,\frac{\sinh(t)\ln\big(\sinh(t)\big)}{\sqrt{1+\sinh^2(t)}}+t+C\\[1mm] =&\,\frac{x\ln(x)}{\sqrt{1+x^2}}+\sinh^{-1}(x)+C \end{align*}

 

 解2

 

x=\mfrac{1}{u} , \dx=-\mfrac{\du}{u^2},則

    \begin{align*} &\,\int\frac{u\ln(u)}{(1+u^2)^{\frac{3}{2}}}\du\\[1mm] =&\,-\frac{\ln(u)}{\sqrt{1+u^2}} +\int\frac{\du}{u\sqrt{1+u^2}}\\[1mm] =&\,\frac{x\ln(x)}{\sqrt{1+x^2}} -\int\frac{\dx}{\sqrt{1+x^2}}\\[1mm] =&\,\frac{x\ln(x)}{\sqrt{1+x^2}}+\sinh^{-1}(x)+C \end{align*}

 


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