$\displaystyle
\int\limits_{ - 2}^2 {\frac{{dx}}
{{4 + x^2 }} = \int\limits_{\frac{{ - \pi }}
{4}}^{\frac{\pi }
{4}} {\frac{{2\sec ^2 u}}
{{4\sec ^2 u}}du} }
$
My question is how do I go from 2,-2 to pi/4, -pi/4? Or, how do I get pi/4?
Thanks.
$\displaystyle
\int\limits_{ - 2}^2 {\frac{{dx}}
{{4 + x^2 }} = \int\limits_{\frac{{ - \pi }}
{4}}^{\frac{\pi }
{4}} {\frac{{2\sec ^2 u}}
{{4\sec ^2 u}}du} }
$
My question is how do I go from 2,-2 to pi/4, -pi/4? Or, how do I get pi/4?
Thanks.
You've made the substitution $\displaystyle x = 2 \tan u$.
$\displaystyle x = 2 \Rightarrow 2 = 2 \tan u \Rightarrow \tan u = 1 \Rightarrow u = \frac{\pi}{4}$.
$\displaystyle x = -2 \Rightarrow -2 = 2 \tan u \Rightarrow \tan u = -1 \Rightarrow u = -\frac{\pi}{4}$.