$$$- \frac{a^{2}}{\sin{\left(x \right)}} + \sin{\left(x \right)}$$$$$$x$$$ 的積分

此計算器會求出 $$$- \frac{a^{2}}{\sin{\left(x \right)}} + \sin{\left(x \right)}$$$$$$x$$$ 的不定積分/原函數,並顯示步驟。

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您的輸入

$$$\int \left(- \frac{a^{2}}{\sin{\left(x \right)}} + \sin{\left(x \right)}\right)\, dx$$$

解答

逐項積分:

$${\color{red}{\int{\left(- \frac{a^{2}}{\sin{\left(x \right)}} + \sin{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{a^{2}}{\sin{\left(x \right)}} d x} + \int{\sin{\left(x \right)} d x}\right)}}$$

使用倍角公式 $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$ 重寫正弦:

$$\int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{a^{2}}{\sin{\left(x \right)}} d x}}} = \int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{a^{2}}{2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}} d x}}}$$

將分子與分母同時乘以 $$$\sec^2\left(\frac{x}{2} \right)$$$:

$$\int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{a^{2}}{2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}} d x}}} = \int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{a^{2} \sec^{2}{\left(\frac{x}{2} \right)}}{2 \tan{\left(\frac{x}{2} \right)}} d x}}}$$

$$$u=\tan{\left(\frac{x}{2} \right)}$$$

$$$du=\left(\tan{\left(\frac{x}{2} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} dx$$$ (步驟見»),並可得 $$$\sec^{2}{\left(\frac{x}{2} \right)} dx = 2 du$$$

該積分可改寫為

$$\int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{a^{2} \sec^{2}{\left(\frac{x}{2} \right)}}{2 \tan{\left(\frac{x}{2} \right)}} d x}}} = \int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{a^{2}}{u} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=a^{2}$$$$$$f{\left(u \right)} = \frac{1}{u}$$$

$$\int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{a^{2}}{u} d u}}} = \int{\sin{\left(x \right)} d x} - {\color{red}{a^{2} \int{\frac{1}{u} d u}}}$$

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

$$- a^{2} {\color{red}{\int{\frac{1}{u} d u}}} + \int{\sin{\left(x \right)} d x} = - a^{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}} + \int{\sin{\left(x \right)} d x}$$

回顧一下 $$$u=\tan{\left(\frac{x}{2} \right)}$$$

$$- a^{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\sin{\left(x \right)} d x} = - a^{2} \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}\right| \right)} + \int{\sin{\left(x \right)} d x}$$

正弦函數的積分為 $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$

$$- a^{2} \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} + {\color{red}{\int{\sin{\left(x \right)} d x}}} = - a^{2} \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} + {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$

因此,

$$\int{\left(- \frac{a^{2}}{\sin{\left(x \right)}} + \sin{\left(x \right)}\right)d x} = - a^{2} \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} - \cos{\left(x \right)}$$

加上積分常數:

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

答案

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


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