$$$\tan^{2}{\left(x \right)}$$$'nin integrali

Bulun: $$$\int \tan^{2}{\left(x \right)}\, dx$$$.

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

O halde $$$x=\operatorname{atan}{\left(u \right)}$$$ ve $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (adımlar » görülebilir).

O halde,

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

Kesri yeniden yazın ve parçalara ayırın:

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

Her terimin integralini alın:

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

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

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

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

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

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

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

Dolayısıyla,

$$\int{\tan^{2}{\left(x \right)} d x} = \tan{\left(x \right)} - \operatorname{atan}{\left(\tan{\left(x \right)} \right)}$$

Sadeleştirin:

$$\int{\tan^{2}{\left(x \right)} d x} = - x + \tan{\left(x \right)}$$

İntegrasyon sabitini ekleyin:

$$\int{\tan^{2}{\left(x \right)} d x} = - x + \tan{\left(x \right)}+C$$

$$$\int \tan^{2}{\left(x \right)}\, dx = \left(- x + \tan{\left(x \right)}\right) + C$$$A