$$$\frac{1}{a^{2} \cos^{2}{\left(x \right)} + b^{2} \sin^{2}{\left(x \right)}}$$$$$$x$$$ 的積分

此計算器會求出 $$$\frac{1}{a^{2} \cos^{2}{\left(x \right)} + b^{2} \sin^{2}{\left(x \right)}}$$$$$$x$$$ 的不定積分/原函數,並顯示步驟。

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您的輸入

$$$\int \frac{1}{a^{2} \cos^{2}{\left(x \right)} + b^{2} \sin^{2}{\left(x \right)}}\, dx$$$

解答

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

$${\color{red}{\int{\frac{1}{a^{2} \cos^{2}{\left(x \right)} + b^{2} \sin^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{a^{2} + b^{2} \tan^{2}{\left(x \right)}} d x}}}$$

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

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

因此,

$${\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{a^{2} + b^{2} \tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{a^{2} + b^{2} u^{2}} d u}}}$$

$$$v=\frac{u \left|{b}\right|}{\left|{a}\right|}$$$

$$$dv=\left(\frac{u \left|{b}\right|}{\left|{a}\right|}\right)^{\prime }du = \frac{\left|{b}\right|}{\left|{a}\right|} du$$$ (步驟見»),並可得 $$$du = \frac{\left|{a}\right| dv}{\left|{b}\right|}$$$

該積分可改寫為

$${\color{red}{\int{\frac{1}{a^{2} + b^{2} u^{2}} d u}}} = {\color{red}{\int{\frac{\left|{a}\right|}{a^{2} \left(v^{2} + 1\right) \left|{b}\right|} d v}}}$$

套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$,使用 $$$c=\frac{\left|{a}\right|}{a^{2} \left|{b}\right|}$$$$$$f{\left(v \right)} = \frac{1}{v^{2} + 1}$$$

$${\color{red}{\int{\frac{\left|{a}\right|}{a^{2} \left(v^{2} + 1\right) \left|{b}\right|} d v}}} = {\color{red}{\frac{\left|{a}\right| \int{\frac{1}{v^{2} + 1} d v}}{a^{2} \left|{b}\right|}}}$$

$$$\frac{1}{v^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$

$$\frac{\left|{a}\right| {\color{red}{\int{\frac{1}{v^{2} + 1} d v}}}}{a^{2} \left|{b}\right|} = \frac{\left|{a}\right| {\color{red}{\operatorname{atan}{\left(v \right)}}}}{a^{2} \left|{b}\right|}$$

回顧一下 $$$v=\frac{u \left|{b}\right|}{\left|{a}\right|}$$$

$$\frac{\left|{a}\right| \operatorname{atan}{\left({\color{red}{v}} \right)}}{a^{2} \left|{b}\right|} = \frac{\left|{a}\right| \operatorname{atan}{\left({\color{red}{\frac{u \left|{b}\right|}{\left|{a}\right|}}} \right)}}{a^{2} \left|{b}\right|}$$

回顧一下 $$$u=\tan{\left(x \right)}$$$

$$\frac{\left|{a}\right| \operatorname{atan}{\left(\frac{\left|{b}\right| {\color{red}{u}}}{\left|{a}\right|} \right)}}{a^{2} \left|{b}\right|} = \frac{\left|{a}\right| \operatorname{atan}{\left(\frac{\left|{b}\right| {\color{red}{\tan{\left(x \right)}}}}{\left|{a}\right|} \right)}}{a^{2} \left|{b}\right|}$$

因此,

$$\int{\frac{1}{a^{2} \cos^{2}{\left(x \right)} + b^{2} \sin^{2}{\left(x \right)}} d x} = \frac{\left|{a}\right| \operatorname{atan}{\left(\frac{\tan{\left(x \right)} \left|{b}\right|}{\left|{a}\right|} \right)}}{a^{2} \left|{b}\right|}$$

化簡:

$$\int{\frac{1}{a^{2} \cos^{2}{\left(x \right)} + b^{2} \sin^{2}{\left(x \right)}} d x} = \frac{\left|{\frac{a}{b}}\right| \operatorname{atan}{\left(\tan{\left(x \right)} \left|{\frac{b}{a}}\right| \right)}}{a^{2}}$$

加上積分常數:

$$\int{\frac{1}{a^{2} \cos^{2}{\left(x \right)} + b^{2} \sin^{2}{\left(x \right)}} d x} = \frac{\left|{\frac{a}{b}}\right| \operatorname{atan}{\left(\tan{\left(x \right)} \left|{\frac{b}{a}}\right| \right)}}{a^{2}}+C$$

答案

$$$\int \frac{1}{a^{2} \cos^{2}{\left(x \right)} + b^{2} \sin^{2}{\left(x \right)}}\, dx = \frac{\left|{\frac{a}{b}}\right| \operatorname{atan}{\left(\tan{\left(x \right)} \left|{\frac{b}{a}}\right| \right)}}{a^{2}} + C$$$A


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