# Evaluate the integral int \ sinx/(2+cos^2x) \ dx ?

Jan 14, 2018

$\int \setminus \sin \frac{x}{2 + {\cos}^{2} x} \setminus \mathrm{dx} = - \frac{\sqrt{2}}{2} \arctan \left(\cos \frac{x}{\sqrt{2}}\right) + C$

#### Explanation:

We seek:

$I = \int \setminus \sin \frac{x}{2 + {\cos}^{2} x} \setminus \mathrm{dx}$

If we look at the denominator then a substitution:

$2 {u}^{2} = {\cos}^{2} x$

looks promising, so let us try the substitution:

$u = \cos \frac{x}{\sqrt{2}} \implies 2 {u}^{2} = {\cos}^{2} x$

Differentiating implicitly wrt $x$:

$\frac{\mathrm{du}}{\mathrm{dx}} = - \sin \frac{x}{\sqrt{2}}$

So substituting into the integral, it becomes:

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

This is now a standard result, thus:

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

Then restoring the substitution:

$I = - \frac{\sqrt{2}}{2} \arctan \left(\cos \frac{x}{\sqrt{2}}\right) + C$