$$$\sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right)$$$'nin integrali

Bulun: $$$\int \sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right)\, dx$$$.

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\sqrt{2}$$$ ve $$$f{\left(x \right)} = \cot^{2}{\left(x \right)} - 1$$$ ile uygula:

$${\color{red}{\int{\sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right) d x}}} = {\color{red}{\sqrt{2} \int{\left(\cot^{2}{\left(x \right)} - 1\right)d x}}}$$

Her terimin integralini alın:

$$\sqrt{2} {\color{red}{\int{\left(\cot^{2}{\left(x \right)} - 1\right)d x}}} = \sqrt{2} {\color{red}{\left(- \int{1 d x} + \int{\cot^{2}{\left(x \right)} d x}\right)}}$$

$$$c=1$$$ kullanarak $$$\int c\, dx = c x$$$ sabit kuralını uygula:

$$\sqrt{2} \left(\int{\cot^{2}{\left(x \right)} d x} - {\color{red}{\int{1 d x}}}\right) = \sqrt{2} \left(\int{\cot^{2}{\left(x \right)} d x} - {\color{red}{x}}\right)$$

$$$u=\cot{\left(x \right)}$$$ olsun.

Böylece $$$du=\left(\cot{\left(x \right)}\right)^{\prime }dx = - \csc^{2}{\left(x \right)} dx$$$ (adımlar » görülebilir) ve $$$\csc^{2}{\left(x \right)} dx = - du$$$ elde ederiz.

O halde,

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

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=-1$$$ ve $$$f{\left(u \right)} = \frac{u^{2}}{u^{2} + 1}$$$ ile uygula:

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

Kesri yeniden yazın ve parçalara ayırın:

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

Her terimin integralini alın:

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

$$$c=1$$$ kullanarak $$$\int c\, du = c u$$$ sabit kuralını uygula:

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

$$$\frac{1}{u^{2} + 1}$$$'nin integrali $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

Hatırlayın ki $$$u=\cot{\left(x \right)}$$$:

$$\sqrt{2} \left(- x + \operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}}\right) = \sqrt{2} \left(- x + \operatorname{atan}{\left({\color{red}{\cot{\left(x \right)}}} \right)} - {\color{red}{\cot{\left(x \right)}}}\right)$$

Dolayısıyla,

$$\int{\sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right) d x} = \sqrt{2} \left(- x - \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right)$$

İntegrasyon sabitini ekleyin:

$$\int{\sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right) d x} = \sqrt{2} \left(- x - \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right)+C$$

$$$\int \sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right)\, dx = \sqrt{2} \left(- x - \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) + C$$$A