$$$- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2$$$ 的積分

此計算器將求出 $$$- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2$$$ 的不定積分(原函數),並顯示步驟。

相關計算器: 定積分與廣義積分計算器

請不要使用任何微分符號,例如 $$$dx$$$$$$dy$$$ 等。
留空以自動偵測。

如果計算器未能計算某些內容,或您發現了錯誤,或您有任何建議/回饋,請聯絡我們

您的輸入

$$$\int \left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)\, dx$$$

解答

逐項積分:

$${\color{red}{\int{\left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)d x}}} = {\color{red}{\left(\int{2 d x} - \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}\right)}}$$

配合 $$$c=2$$$,應用常數法則 $$$\int c\, dx = c x$$$

$$- \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} + {\color{red}{\int{2 d x}}} = - \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} + {\color{red}{\left(2 x\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=3$$$$$$f{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}}$$$

$$2 x - {\color{red}{\int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}} = 2 x - {\color{red}{\left(3 \int{\frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}\right)}}$$

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

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

所以,

$$2 x - 3 {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}} = 2 x - 3 {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}}$$

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

$$2 x - 3 {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}} = 2 x - 3 {\color{red}{\left(- \int{\frac{1}{u^{2}} d u}\right)}}$$

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

$$2 x + 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=2 x + 3 {\color{red}{\int{u^{-2} d u}}}=2 x + 3 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=2 x + 3 {\color{red}{\left(- u^{-1}\right)}}=2 x + 3 {\color{red}{\left(- \frac{1}{u}\right)}}$$

回顧一下 $$$u=\cos{\left(x \right)}$$$

$$2 x - 3 {\color{red}{u}}^{-1} = 2 x - 3 {\color{red}{\cos{\left(x \right)}}}^{-1}$$

因此,

$$\int{\left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)d x} = 2 x - \frac{3}{\cos{\left(x \right)}}$$

加上積分常數:

$$\int{\left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)d x} = 2 x - \frac{3}{\cos{\left(x \right)}}+C$$

答案

$$$\int \left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)\, dx = \left(2 x - \frac{3}{\cos{\left(x \right)}}\right) + C$$$A