$$$\sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)}$$$'nin integrali

Bulun: $$$\int \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$$.

Integrand ifadesini, $$$\alpha=x$$$ ile $$$\sin^2\left( \alpha \right)=\frac{1}{2}-\frac{1}{2}\cos\left(2 \alpha \right)-$$$ ve $$$\beta=x$$$ ile $$$\cos^2\left( \beta \right)=\frac{1}{2}+\frac{1}{2}\cos\left(2 \beta \right)+$$$ kuvvet indirgeme formüllerini kullanarak yeniden yazın:

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

İfadeyi genişletin:

$${\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2} d x}}} = {\color{red}{\int{\left(- \frac{\cos^{3}{\left(2 x \right)}}{8} - \frac{\cos^{2}{\left(2 x \right)}}{8} + \frac{\cos{\left(2 x \right)}}{8} + \frac{1}{8}\right)d x}}}$$

Her terimin integralini alın:

$${\color{red}{\int{\left(- \frac{\cos^{3}{\left(2 x \right)}}{8} - \frac{\cos^{2}{\left(2 x \right)}}{8} + \frac{\cos{\left(2 x \right)}}{8} + \frac{1}{8}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{8} d x} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{2}{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x}\right)}}$$

$$$c=\frac{1}{8}$$$ kullanarak $$$\int c\, dx = c x$$$ sabit kuralını uygula:

$$\int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{2}{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} + {\color{red}{\int{\frac{1}{8} d x}}} = \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{2}{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} + {\color{red}{\left(\frac{x}{8}\right)}}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{8}$$$ ve $$$f{\left(x \right)} = \cos^{2}{\left(2 x \right)}$$$ ile uygula:

$$\frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - {\color{red}{\int{\frac{\cos^{2}{\left(2 x \right)}}{8} d x}}} = \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - {\color{red}{\left(\frac{\int{\cos^{2}{\left(2 x \right)} d x}}{8}\right)}}$$

$$$u=2 x$$$ olsun.

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

Dolayısıyla,

$$\frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos^{2}{\left(2 x \right)} d x}}}}{8} = \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{2} d u}}}}{8}$$

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

$$\frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{2} d u}}}}{8} = \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\frac{\int{\cos^{2}{\left(u \right)} d u}}{2}\right)}}}{8}$$

Kuvvet indirgeme formülü $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$'i $$$\alpha= u $$$ ile uygula:

$$\frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}}{16} = \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{16}$$

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

$$\frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{16} = \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}}{16}$$

Her terimin integralini alın:

$$\frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{32} = \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{32}$$

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

$$\frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\int{\cos{\left(2 u \right)} d u}}{32} - \frac{{\color{red}{\int{1 d u}}}}{32} = \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\int{\cos{\left(2 u \right)} d u}}{32} - \frac{{\color{red}{u}}}{32}$$

$$$v=2 u$$$ olsun.

Böylece $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (adımlar » görülebilir) ve $$$du = \frac{dv}{2}$$$ elde ederiz.

İntegral şu hale gelir

$$- \frac{u}{32} + \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{32} = - \frac{u}{32} + \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32}$$

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

$$- \frac{u}{32} + \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32} = - \frac{u}{32} + \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{32}$$

Kosinüsün integrali $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$- \frac{u}{32} + \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = - \frac{u}{32} + \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\sin{\left(v \right)}}}}{64}$$

Hatırlayın ki $$$v=2 u$$$:

$$- \frac{u}{32} + \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\sin{\left({\color{red}{v}} \right)}}{64} = - \frac{u}{32} + \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{64}$$

Hatırlayın ki $$$u=2 x$$$:

$$\frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\sin{\left(2 {\color{red}{u}} \right)}}{64} - \frac{{\color{red}{u}}}{32} = \frac{x}{8} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\sin{\left(2 {\color{red}{\left(2 x\right)}} \right)}}{64} - \frac{{\color{red}{\left(2 x\right)}}}{32}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{8}$$$ ve $$$f{\left(x \right)} = \cos^{3}{\left(2 x \right)}$$$ ile uygula:

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - {\color{red}{\int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x}}} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - {\color{red}{\left(\frac{\int{\cos^{3}{\left(2 x \right)} d x}}{8}\right)}}$$

$$$u=2 x$$$ olsun.

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

Dolayısıyla,

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos^{3}{\left(2 x \right)} d x}}}}{8} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos^{3}{\left(u \right)}}{2} d u}}}}{8}$$

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

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos^{3}{\left(u \right)}}{2} d u}}}}{8} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\frac{\int{\cos^{3}{\left(u \right)} d u}}{2}\right)}}}{8}$$

Bir kosinüsü ayırın ve geri kalan her şeyi sinüs cinsinden yazın; $$$\alpha= u $$$ ile $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ formülünü kullanarak.:

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos^{3}{\left(u \right)} d u}}}}{16} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)} d u}}}}{16}$$

$$$v=\sin{\left(u \right)}$$$ olsun.

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

İntegral şu şekilde yeniden yazılabilir:

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)} d u}}}}{16} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(1 - v^{2}\right)d v}}}}{16}$$

Her terimin integralini alın:

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(1 - v^{2}\right)d v}}}}{16} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\int{1 d v} - \int{v^{2} d v}\right)}}}{16}$$

$$$c=1$$$ kullanarak $$$\int c\, dv = c v$$$ sabit kuralını uygula:

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \frac{\int{v^{2} d v}}{16} - \frac{{\color{red}{\int{1 d v}}}}{16} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \frac{\int{v^{2} d v}}{16} - \frac{{\color{red}{v}}}{16}$$

Kuvvet kuralını $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=2$$$ ile uygulayın:

$$- \frac{v}{16} + \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \frac{{\color{red}{\int{v^{2} d v}}}}{16}=- \frac{v}{16} + \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \frac{{\color{red}{\frac{v^{1 + 2}}{1 + 2}}}}{16}=- \frac{v}{16} + \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \frac{{\color{red}{\left(\frac{v^{3}}{3}\right)}}}{16}$$

Hatırlayın ki $$$v=\sin{\left(u \right)}$$$:

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{v}}}{16} + \frac{{\color{red}{v}}^{3}}{48} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\sin{\left(u \right)}}}}{16} + \frac{{\color{red}{\sin{\left(u \right)}}}^{3}}{48}$$

Hatırlayın ki $$$u=2 x$$$:

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{\sin{\left({\color{red}{u}} \right)}}{16} + \frac{\sin^{3}{\left({\color{red}{u}} \right)}}{48} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{16} + \frac{\sin^{3}{\left({\color{red}{\left(2 x\right)}} \right)}}{48}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{8}$$$ ve $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$ ile uygula:

$$\frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + {\color{red}{\int{\frac{\cos{\left(2 x \right)}}{8} d x}}} = \frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + {\color{red}{\left(\frac{\int{\cos{\left(2 x \right)} d x}}{8}\right)}}$$

İntegral $$$\int{\cos{\left(2 x \right)} d x}$$$ zaten hesaplandı:

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

Dolayısıyla,

$$\frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{8} = \frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\left(\frac{\sin{\left(2 x \right)}}{2}\right)}}}{8}$$

Dolayısıyla,

$$\int{\sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)} d x} = \frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}$$

İntegrasyon sabitini ekleyin:

$$\int{\sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)} d x} = \frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}+C$$

$$$\int \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = \left(\frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}\right) + C$$$A