# How do I integrate f(x)=sec(x)/(4-3tan(x))?

Apr 22, 2015

$\int \frac{\sec \left(x\right)}{4 - 3 \tan \left(x\right)} \mathrm{dx} = \int \frac{\frac{1}{\cos} \left(x\right)}{4 - 3 \tan \left(x\right)} \mathrm{dx}$

Let's start with $t = \tan \left(\frac{x}{2}\right)$

$\sec \left(x\right) = \frac{1 + {t}^{2}}{1 - {t}^{2}}$

$\tan \left(x\right) = \frac{2 t}{1 - {t}^{2}}$

$\mathrm{dx} = \frac{2}{1 + {t}^{2}} \mathrm{dt}$

So, the integral become :

$\implies \int \frac{\frac{1 + {t}^{2}}{1 - {t}^{2}}}{4 - \frac{6 t}{1 - {t}^{2}}} \cdot \frac{2}{1 + {t}^{2}} \mathrm{dt}$

$\implies \int \frac{\frac{1 + {t}^{2}}{1 - {t}^{2}}}{\frac{4 \left(1 - {t}^{2}\right)}{1 - {t}^{2}} - \frac{6 t}{1 - {t}^{2}}} \cdot \frac{2}{1 + {t}^{2}} \mathrm{dt}$

$\implies \int \frac{\frac{1 + {t}^{2}}{1 - {t}^{2}}}{\frac{- 4 {t}^{2} - 6 t + 4}{1 - {t}^{2}}} \cdot \frac{2}{1 + {t}^{2}} \mathrm{dt}$

$\implies \int \frac{1 + {t}^{2}}{\left(- 4 {t}^{2} - 6 t + 4\right)} \cdot \frac{2}{1 + {t}^{2}} \mathrm{dt}$

$\implies \int \frac{2}{2 \left(- 2 {t}^{2} - 3 t + 2\right)} \mathrm{dt}$

$\implies \int \frac{1}{- 2 {t}^{2} - 3 t + 2} \mathrm{dt}$

$\Delta = {b}^{2} - 4 a c = 25$

${x}_{1} = \frac{1}{2}$

${x}_{2} = - 2$

Factorize :

$\implies \int - \frac{1}{\left(2 t - 1\right) \left(t + 2\right)} \mathrm{dt}$

Partial fraction :

$\implies \frac{1}{5} \int \frac{1}{t + 2} \mathrm{dt} - \frac{1}{5} \int \frac{2}{2 t - 1} \mathrm{dt}$

$\implies \frac{1}{5} \left[\ln \left(| t + 2 |\right)\right] - \frac{1}{5} \left[\ln \left(| 2 t - 1 |\right)\right] + C$

=> 1/5((ln(|t+2|)-ln(|2t-1|))+C

Substitute back for $t = \tan \left(\frac{1}{2} x\right)$

$\implies \frac{1}{5} \left(\ln \left(| \tan \left(\frac{1}{2} x\right) + 2 |\right) - \ln \left(| 2 \tan \left(\frac{1}{2} x\right) - 1 |\right)\right) + C$