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

Jan 9, 2017

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

#### Explanation:

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

We will make the substitution $\cos x = \tan \theta$. Differentiating this shows that $- \sin x \textcolor{w h i t e}{.} \mathrm{dx} = {\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta$.

When making this substitution from $x$ to $\theta$, we also need to change the bounds.

• $x = \frac{\pi}{2} \implies \cos \left(x\right) = \cos \left(\frac{\pi}{2}\right) = 0 = \tan \left(\theta\right) \implies \theta = 0$
• $x = \pi \implies \cos \left(x\right) = \cos \left(\pi\right) = - 1 = \tan \left(\theta\right) \implies \theta = - \frac{\pi}{4}$

So:

$I = - {\int}_{\pi / 2}^{\pi} \frac{- \sin x \textcolor{w h i t e}{.} \mathrm{dx}}{1 + {\cos}^{2} x} = - {\int}_{0}^{- \pi / 4} \frac{{\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta}{1 + {\tan}^{2} \theta}$

Reversing the order of the bounds with the negative sign and using the identity ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$:

$I = {\int}_{- \pi / 4}^{0} d \theta = {\left[\theta\right]}_{- \pi / 4}^{0} = 0 - \left(- \frac{\pi}{4}\right) = \frac{\pi}{4}$