$$$x \sec^{2}{\left(x \right)}$$$ 的積分

此計算器將求出 $$$x \sec^{2}{\left(x \right)}$$$ 的不定積分(原函數),並顯示步驟。

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您的輸入

$$$\int x \sec^{2}{\left(x \right)}\, dx$$$

解答

對於積分 $$$\int{x \sec^{2}{\left(x \right)} d x}$$$,使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$

$$$\operatorname{u}=x$$$$$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$

$$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$(步驟見 »),且 $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$(步驟見 »)。

因此,

$${\color{red}{\int{x \sec^{2}{\left(x \right)} d x}}}={\color{red}{\left(x \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot 1 d x}\right)}}={\color{red}{\left(x \tan{\left(x \right)} - \int{\tan{\left(x \right)} d x}\right)}}$$

將切線改寫為 $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

$$x \tan{\left(x \right)} - {\color{red}{\int{\tan{\left(x \right)} d x}}} = x \tan{\left(x \right)} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

$$$u=\cos{\left(x \right)}$$$

$$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (步驟見»),並可得 $$$\sin{\left(x \right)} dx = - du$$$

該積分可改寫為

$$x \tan{\left(x \right)} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = x \tan{\left(x \right)} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=-1$$$$$$f{\left(u \right)} = \frac{1}{u}$$$

$$x \tan{\left(x \right)} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = x \tan{\left(x \right)} - {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$

$$x \tan{\left(x \right)} + {\color{red}{\int{\frac{1}{u} d u}}} = x \tan{\left(x \right)} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回顧一下 $$$u=\cos{\left(x \right)}$$$

$$x \tan{\left(x \right)} + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = x \tan{\left(x \right)} + \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)}$$

因此,

$$\int{x \sec^{2}{\left(x \right)} d x} = x \tan{\left(x \right)} + \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}$$

加上積分常數:

$$\int{x \sec^{2}{\left(x \right)} d x} = x \tan{\left(x \right)} + \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}+C$$

答案

$$$\int x \sec^{2}{\left(x \right)}\, dx = \left(x \tan{\left(x \right)} + \ln\left(\left|{\cos{\left(x \right)}}\right|\right)\right) + C$$$A


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