幾道積分練習題

積分題 1

\begin{align*}
\int\frac{\sin(x)\cos(x)}{\sin(x)+\cos(x)}\mathrm{d}x
\end{align*}

解 1

\begin{align}
&\,\int\frac{\sin(x)\cos(x)}{\sin(x)+\cos(x)}\mathrm{d}x\\[3mm]
=&\,\frac{1}{\sqrt{2}}\int\frac{\sin(x)\cos(x)}{\sin(x+\frac{\pi}{4})}\mathrm{d}x\\[3mm]
=&\,\frac{1}{2\sqrt{2}}\frac{\Big[\sin(u)-\cos(u)\Big]\Big[\cos(u)+\sin(u)\Big]}{\sin(u)}\mathrm{d}u\\[3mm]
=&\,\frac{1}{2\sqrt{2}}\int\frac{2\sin^2(u)-1}{\sin(u)}\du\\[3mm]
=&\,\frac{1}{\sqrt{2}}\int\sin(u)\mathrm{d}u
-\frac{1}{2\sqrt{2}}\int\csc(u)\mathrm{d}u\\[3mm]
=&\,-\frac{\cos(x+\frac{\pi}{4})}{\sqrt{2}}\\[3mm]
&\,\;-\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)}\mathrm{d}x\\[3mm]
=&\,\frac{1}{2}\int\frac{\big(\sin(x)+\cos(x)\big)^2-1}{\sin(x)+\cos(x)}\mathrm{d}x\\[3mm]
=&\,\frac{1}{2}\int\big(\sin(x)+\cos(x)\big)\mathrm{d}x\\[3mm]
&\,\quad\frac{1}{2\sqrt{2}}\int\frac{1}{\sin(x+\frac{\pi}{4})}\mathrm{d}x\\[3mm]
=&\,\frac{1}{2}\big(-\cos(x)+\sin(x)\big)\\[3mm]
&\,\quad-\frac{1}{2\sqrt{2}}\ln\abs{\csc(x+\frac{\pi}{4})
-\cot(x+\frac{\pi}{4})}+C
\end{align}

積分題 2

\begin{align*}
\int e^{\sin(x)}\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}\mathrm{d}x
\end{align*}

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

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

於是

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

積分題 3

\int\frac{\ln(x)}{(1+x^2)^{\frac{3}{2}}}\mathrm{d}x

解 1

設 \(x=\sinh(t) , \mathrm{d}x=\cosh(t)\mathrm{d}t\) ，則

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

解 2

設 \(x=\frac{1}{u} , \mathrm{d}x=-\frac{\mathrm{d}u}{u^2}\)，則

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