How do you use substitution to integrate sec(v + pi/2) * tan(v + pi/2)?

Sep 22, 2015

See the explanation.

Explanation:

$t = \frac{1}{\sec} \left(v + \frac{\pi}{2}\right)$

$\mathrm{dt} = \left({\left(\sec \left(v + \frac{\pi}{2}\right)\right)}^{-} 1\right) ' \mathrm{dv} =$

$\mathrm{dt} = - \frac{1}{\sec} ^ 2 \left(v + \frac{\pi}{2}\right) \sec \left(v + \frac{\pi}{2}\right) \tan \left(v + \frac{\pi}{2}\right) \mathrm{dv}$

$\mathrm{dt} = - \frac{\tan \left(v + \frac{\pi}{2}\right) \mathrm{dv}}{\sec} \left(v + \frac{\pi}{2}\right) = - t \tan \left(v + \frac{\pi}{2}\right) \mathrm{dv}$

$\tan \left(v + \frac{\pi}{2}\right) \mathrm{dv} = - \frac{\mathrm{dt}}{t}$

$\int \sec \left(v + \frac{\pi}{2}\right) \tan \left(v + \frac{\pi}{2}\right) \mathrm{dv} = \int \frac{1}{t} \left(- \frac{\mathrm{dt}}{t}\right) =$

$= - \int {t}^{-} 2 \mathrm{dt} = \frac{1}{t} + C = \sec \left(v + \frac{\pi}{2}\right) + C$