$$$\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}$$$ 的積分

此計算器將求出 $$$\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}$$$ 的不定積分(原函數),並顯示步驟。

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您的輸入

$$$\int \frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}\, dx$$$

解答

將分子與分母同時乘以 $$$\frac{1}{\cosh^{2}{\left(x \right)}}$$$,並將 $$$\frac{\cosh^{2}{\left(x \right)}}{\sinh^{2}{\left(x \right)}}$$$ 轉換為 $$$\frac{1}{\tanh^{2}{\left(x \right)}}$$$:

$${\color{red}{\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\cosh^{4}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}}$$

取兩個雙曲餘弦,並使用公式 $$$\cosh^{2}{\left(x \right)}=\frac{1}{1 - \tanh^{2}{\left(x \right)}}$$$ 將其他雙曲餘弦改寫為以雙曲正切表示。:

$${\color{red}{\int{\frac{1}{\cosh^{4}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1 - \tanh^{2}{\left(x \right)}}{\cosh^{2}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}}$$

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

$$$du=\left(\tanh{\left(x \right)}\right)^{\prime }dx = \operatorname{sech}^{2}{\left(x \right)} dx$$$ (步驟見»),並可得 $$$\operatorname{sech}^{2}{\left(x \right)} dx = du$$$

該積分變為

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

Expand the expression:

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

逐項積分:

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

配合 $$$c=1$$$,應用常數法則 $$$\int c\, du = c u$$$

$$\int{\frac{1}{u^{2}} d u} - {\color{red}{\int{1 d u}}} = \int{\frac{1}{u^{2}} d u} - {\color{red}{u}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=-2$$$

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

回顧一下 $$$u=\tanh{\left(x \right)}$$$

$$- {\color{red}{u}}^{-1} - {\color{red}{u}} = - {\color{red}{\tanh{\left(x \right)}}}^{-1} - {\color{red}{\tanh{\left(x \right)}}}$$

因此,

$$\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x} = - \tanh{\left(x \right)} - \frac{1}{\tanh{\left(x \right)}}$$

加上積分常數:

$$\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x} = - \tanh{\left(x \right)} - \frac{1}{\tanh{\left(x \right)}}+C$$

答案

$$$\int \frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}\, dx = \left(- \tanh{\left(x \right)} - \frac{1}{\tanh{\left(x \right)}}\right) + C$$$A