# Evaluate the integral  int 1/(5+3cosx) dx ?

Jun 23, 2017

$\int \setminus \frac{1}{5 + 3 \cos x} \setminus \mathrm{dx} = \frac{1}{2} \arctan \left(\frac{1}{2} \tan \left(\frac{x}{2}\right)\right) + C$

#### Explanation:

We want to evaluate:

$I = \int \setminus \frac{1}{5 + 3 \cos x} \setminus \mathrm{dx}$

If we use the trigonometry half angle tangent formula then we have:

$\cos \alpha = \frac{1 - {\tan}^{2} \left(\frac{\alpha}{2}\right)}{1 + {\tan}^{2} \left(\frac{\alpha}{2}\right)}$

Let us perform the requested change of variable via the substation:

$\tan \left(\frac{x}{2}\right) = u$

Differentiating wrt $x$ and using the identity $1 + {\tan}^{2} \alpha = {\sec}^{2} \alpha$ we get:

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2} {\sec}^{2} \left(\frac{u}{2}\right)$
$\text{ } = \frac{1}{2} \left(1 + {\tan}^{2} \left(\frac{u}{2}\right)\right)$
$\text{ } = \frac{1}{2} \left(1 + {u}^{2}\right)$

$\therefore \frac{2}{1 + {u}^{2}} \setminus \frac{\mathrm{du}}{\mathrm{dx}} = 1$

Then substituting into the integral, we get:

$I = \int \setminus \frac{1}{5 + 3 \setminus \frac{1 - {u}^{2}}{1 + {u}^{2}}} \setminus \frac{2}{1 + {u}^{2}} \setminus \mathrm{du}$

$\setminus \setminus = \int \setminus \frac{1}{\frac{5 \left(1 + {u}^{2}\right) + 3 \left(1 - {u}^{2}\right)}{1 + {u}^{2}}} \setminus \frac{2}{1 + {u}^{2}} \setminus \mathrm{du}$

$\setminus \setminus = \int \setminus \frac{1 + {u}^{2}}{5 + 5 {u}^{2} + 3 - 3 {u}^{2}} \setminus \frac{2}{1 + {u}^{2}} \setminus \mathrm{du}$

$\setminus \setminus = \int \setminus \frac{2}{8 + 2 {u}^{2}} \setminus \mathrm{du}$

$\setminus \setminus = \int \setminus \frac{1}{{2}^{2} + {u}^{2}} \setminus \mathrm{du}$

The above is a standard integral than we can quote

$I = \frac{1}{2} \arctan \left(\frac{u}{2}\right) + C$

Restoring the substitution we get:

$I = \frac{1}{2} \arctan \left(\frac{\tan \left(\frac{x}{2}\right)}{2}\right) + C$
$\setminus \setminus = \frac{1}{2} \arctan \left(\frac{1}{2} \tan \left(\frac{x}{2}\right)\right) + C$