# How do you evaluate int cosx/(1+sin^2x) from [0, pi/2]?

Nov 9, 2016

$\int \cos \frac{x}{1 + {\sin}^{2} x} \mathrm{dx} = \frac{\pi}{4}$

#### Explanation:

First examining without the bounds:

$I = \int \cos \frac{x}{1 + {\sin}^{2} x} \mathrm{dx}$

Let $\sin x = \tan \theta$. This may look like a wild substitution, but the goal is to get the denominator of the fraction to $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$. Note that differentiating $\sin x = \tan \theta$ gives $\cos x \mathrm{dx} = {\sec}^{2} \theta d \theta$. It's also helpful that $\cos x \mathrm{dx}$ is already present in the integrand. Substituting these in:

$I = \int {\sec}^{2} \frac{\theta}{1 + {\tan}^{2} \theta} d \theta = \int d \theta = \theta + C$

Since $\sin x = \tan \theta$, we see that $\theta = \arctan \left(\sin x\right)$:

$I = \arctan \left(\sin x\right) + C$

${I}_{B} = {\int}_{0}^{\frac{\pi}{2}} \cos \frac{x}{1 + {\sin}^{2} x} \mathrm{dx} = {\left[\arctan \left(\sin x\right)\right]}_{0}^{\frac{\pi}{2}}$
$\textcolor{w h i t e}{{I}_{B}} = \arctan \left(\sin \left(\frac{\pi}{2}\right)\right) - \arctan \left(\sin \left(0\right)\right)$
$\textcolor{w h i t e}{{I}_{B}} = \arctan \left(1\right) - \arctan \left(0\right)$
$\textcolor{w h i t e}{{I}_{B}} = \frac{\pi}{4} - 0$
$\textcolor{w h i t e}{{I}_{B}} = \frac{\pi}{4}$