Question #668e4

Dec 13, 2017

$\int \frac{\left(2 \sin x + \cos x\right) \cdot \mathrm{dx}}{7 \sin x - 5 \cos x}$=$\frac{9}{74} \cdot x + \frac{17}{74} L n \left(7 \sin x - 5 \cos x\right) + C$

Explanation:

I used $2 \sin x + \cos x = a \cdot \left(7 \sin x - 5 \cos x\right) + b \cdot \left(7 \cos x + 5 \sin x\right)$ equation for calculating this integral,

$\left(7 a + 5 b\right) \cdot \sin x + \left(7 b - 5 a\right) \cdot \cos x = 2 \sin x + \cos x$

After equating coefficients, $7 a + 5 b = 2$ and $7 b - 5 a = 1$

After solving these equations, I found $a = \frac{9}{74}$ and $b = \frac{17}{74}$

Thus,

$\int \frac{\left(2 \sin x + \cos x\right) \cdot \mathrm{dx}}{7 \sin x - 5 \cos x}$

=$\frac{9}{74} \cdot \int \frac{\left(7 \sin x - 5 \cos x\right) \cdot \mathrm{dx}}{7 \sin x - 5 \cos x}$+$\frac{17}{74} \cdot \int \frac{\left(7 \cos x + 5 \sin x\right) \cdot \mathrm{dx}}{7 \sin x - 5 \cos x}$

=$\frac{9}{74} \cdot \int \mathrm{dx}$+$\frac{17}{74} \cdot \int \frac{\left(7 \cos x + 5 \sin x\right) \cdot \mathrm{dx}}{7 \sin x - 5 \cos x}$

=$\frac{9}{74} \cdot x + \frac{17}{74} L n \left(7 \sin x - 5 \cos x\right) + C$