$$$\frac{1}{4 \cos{\left(x \right)} + 5}$$$'nin integrali

Bulun: $$$\int \frac{1}{4 \cos{\left(x \right)} + 5}\, dx$$$.

Integrali alınan ifadeyi $$$\cos{\left(x \right)}=\frac{1 - \tan^{2}{\left(\frac{x}{2} \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} + 1}$$$ formülünü kullanarak yeniden yazın:

$${\color{red}{\int{\frac{1}{4 \cos{\left(x \right)} + 5} d x}}} = {\color{red}{\int{\frac{1}{\frac{4 \left(1 - \tan^{2}{\left(\frac{x}{2} \right)}\right)}{\tan^{2}{\left(\frac{x}{2} \right)} + 1} + 5} d x}}}$$

$$$u=\tan{\left(\frac{x}{2} \right)}$$$ olsun.

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

İntegral şu hale gelir

$${\color{red}{\int{\frac{1}{\frac{4 \left(1 - \tan^{2}{\left(\frac{x}{2} \right)}\right)}{\tan^{2}{\left(\frac{x}{2} \right)} + 1} + 5} d x}}} = {\color{red}{\int{\frac{2}{\left(u^{2} + 1\right) \left(\frac{4 \left(1 - u^{2}\right)}{u^{2} + 1} + 5\right)} d u}}}$$

Sadeleştir:

$${\color{red}{\int{\frac{2}{\left(u^{2} + 1\right) \left(\frac{4 \left(1 - u^{2}\right)}{u^{2} + 1} + 5\right)} d u}}} = {\color{red}{\int{\frac{2}{u^{2} + 9} d u}}}$$

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=2$$$ ve $$$f{\left(u \right)} = \frac{1}{u^{2} + 9}$$$ ile uygula:

$${\color{red}{\int{\frac{2}{u^{2} + 9} d u}}} = {\color{red}{\left(2 \int{\frac{1}{u^{2} + 9} d u}\right)}}$$

$$$v=\frac{u}{3}$$$ olsun.

Böylece $$$dv=\left(\frac{u}{3}\right)^{\prime }du = \frac{du}{3}$$$ (adımlar » görülebilir) ve $$$du = 3 dv$$$ elde ederiz.

Dolayısıyla,

$$2 {\color{red}{\int{\frac{1}{u^{2} + 9} d u}}} = 2 {\color{red}{\int{\frac{1}{3 \left(v^{2} + 1\right)} d v}}}$$

Sabit katsayı kuralı $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$'i $$$c=\frac{1}{3}$$$ ve $$$f{\left(v \right)} = \frac{1}{v^{2} + 1}$$$ ile uygula:

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

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

$$\frac{2 {\color{red}{\int{\frac{1}{v^{2} + 1} d v}}}}{3} = \frac{2 {\color{red}{\operatorname{atan}{\left(v \right)}}}}{3}$$

Hatırlayın ki $$$v=\frac{u}{3}$$$:

$$\frac{2 \operatorname{atan}{\left({\color{red}{v}} \right)}}{3} = \frac{2 \operatorname{atan}{\left({\color{red}{\left(\frac{u}{3}\right)}} \right)}}{3}$$

Hatırlayın ki $$$u=\tan{\left(\frac{x}{2} \right)}$$$:

$$\frac{2 \operatorname{atan}{\left(\frac{{\color{red}{u}}}{3} \right)}}{3} = \frac{2 \operatorname{atan}{\left(\frac{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}{3} \right)}}{3}$$

Dolayısıyla,

$$\int{\frac{1}{4 \cos{\left(x \right)} + 5} d x} = \frac{2 \operatorname{atan}{\left(\frac{\tan{\left(\frac{x}{2} \right)}}{3} \right)}}{3}$$

İntegrasyon sabitini ekleyin:

$$\int{\frac{1}{4 \cos{\left(x \right)} + 5} d x} = \frac{2 \operatorname{atan}{\left(\frac{\tan{\left(\frac{x}{2} \right)}}{3} \right)}}{3}+C$$

$$$\int \frac{1}{4 \cos{\left(x \right)} + 5}\, dx = \frac{2 \operatorname{atan}{\left(\frac{\tan{\left(\frac{x}{2} \right)}}{3} \right)}}{3} + C$$$A