$$$\frac{2 \sin{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}}$$$'nin integrali
İlgili hesap makinesi: Belirli ve Uygunsuz İntegral Hesaplayıcı
Girdiniz
Bulun: $$$\int \frac{2 \sin{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}}\, dx$$$.
Çözüm
Payı yeniden yazın ve kesri ayırın:
$${\color{red}{\int{\frac{2 \sin{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}} d x}}} = {\color{red}{\int{\left(1 + \frac{\sin{\left(x \right)} + \cos{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}}\right)d x}}}$$
Her terimin integralini alın:
$${\color{red}{\int{\left(1 + \frac{\sin{\left(x \right)} + \cos{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}}\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{\sin{\left(x \right)} + \cos{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}} d x}\right)}}$$
$$$c=1$$$ kullanarak $$$\int c\, dx = c x$$$ sabit kuralını uygula:
$$\int{\frac{\sin{\left(x \right)} + \cos{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{\sin{\left(x \right)} + \cos{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}} d x} + {\color{red}{x}}$$
$$$u=\sin{\left(x \right)} - \cos{\left(x \right)}$$$ olsun.
Böylece $$$du=\left(\sin{\left(x \right)} - \cos{\left(x \right)}\right)^{\prime }dx = \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) dx$$$ (adımlar » görülebilir) ve $$$\left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) dx = du$$$ elde ederiz.
İntegral şu şekilde yeniden yazılabilir:
$$x + {\color{red}{\int{\frac{\sin{\left(x \right)} + \cos{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}} d x}}} = x + {\color{red}{\int{\frac{1}{u} d u}}}$$
$$$\frac{1}{u}$$$'nin integrali $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$x + {\color{red}{\int{\frac{1}{u} d u}}} = x + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Hatırlayın ki $$$u=\sin{\left(x \right)} - \cos{\left(x \right)}$$$:
$$x + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = x + \ln{\left(\left|{{\color{red}{\left(\sin{\left(x \right)} - \cos{\left(x \right)}\right)}}}\right| \right)}$$
Dolayısıyla,
$$\int{\frac{2 \sin{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}} d x} = x + \ln{\left(\left|{\sin{\left(x \right)} - \cos{\left(x \right)}}\right| \right)}$$
Sadeleştirin:
$$\int{\frac{2 \sin{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}} d x} = x + \ln{\left(\left|{\cos{\left(x + \frac{\pi}{4} \right)}}\right| \right)} + \frac{\ln{\left(2 \right)}}{2}$$
İntegrasyon sabitini ekleyin (ve ifadeden sabit terimi kaldırın):
$$\int{\frac{2 \sin{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}} d x} = x + \ln{\left(\left|{\cos{\left(x + \frac{\pi}{4} \right)}}\right| \right)}+C$$
Cevap
$$$\int \frac{2 \sin{\left(x \right)}}{\sin{\left(x \right)} - \cos{\left(x \right)}}\, dx = \left(x + \ln\left(\left|{\cos{\left(x + \frac{\pi}{4} \right)}}\right|\right)\right) + C$$$A