# How do you integrate sec(x)/(4-3tan(x)) dx?

Mar 28, 2018

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

#### Explanation:

$\int \sec \frac{x}{4 - 3 \tan x} \cdot \mathrm{dx}$

=$\int \frac{\mathrm{dx}}{4 \cos x - 3 \sin x}$

After using $y = \tan \left(\frac{x}{2}\right)$, $\mathrm{dx} = \frac{2 \mathrm{dy}}{{y}^{2} + 1}$, $\cos x = \frac{1 - {y}^{2}}{{y}^{2} + 1}$ and $\sin x = \frac{2 y}{{y}^{2} + 1}$ transforms, this integral became

$\int \frac{\frac{2 \mathrm{dy}}{{y}^{2} + 1}}{4 \cdot \frac{1 - {y}^{2}}{{y}^{2} + 1} - 3 \cdot \frac{2 y}{{y}^{2} + 1}}$

=$\int \frac{\frac{2 \mathrm{dy}}{{y}^{2} + 1}}{\frac{4 - 6 y - 4 {y}^{2}}{{y}^{2} ^ 1}}$

=-$\int \frac{\mathrm{dy}}{2 {y}^{2} + 3 y - 2}$

=-$\int \frac{\mathrm{dy}}{\left(y + 2\right) \cdot \left(2 y - 1\right)}$

=-$\frac{1}{5} \int \frac{5 \mathrm{dy}}{\left(y + 2\right) \cdot \left(2 y - 1\right)}$

=-$\frac{1}{5} \int \frac{\left(2 y + 4\right) \cdot \mathrm{dy}}{\left(y + 2\right) \left(2 y - 1\right)}$+$\frac{1}{5} \int \frac{\left(2 y - 1\right) \cdot \mathrm{dy}}{\left(y + 2\right) \left(2 y - 1\right)}$

=$\frac{1}{5} \int \frac{\mathrm{dy}}{y + 2} - \frac{2}{5} \int \frac{\mathrm{dy}}{2 y - 1}$

=$\frac{1}{5} \ln \left(y + 2\right) - \frac{1}{5} L n \left(2 y - 1\right) + C$

=$\frac{1}{5} \ln \left(\tan \left(\frac{x}{2}\right) + 2\right) - \frac{1}{5} L n \left(2 \tan \left(\frac{x}{2}\right) - 1\right) + C$