# F(t)=sec pi t/4?

Apr 10, 2018

#### Explanation:

Derivative:
$F \left(t\right) = \sec \left(\frac{\pi t}{4}\right)$

$F \left(t\right) = {\left(\cos \left(\frac{\pi t}{4}\right)\right)}^{-} 1$

$F ' \left(t\right) = - {\left(\cos \left(\frac{\pi t}{4}\right)\right)}^{-} 2 \times - \sin \left(\frac{\pi t}{4}\right) \times \frac{\pi}{4}$

$F ' \left(t\right) = \frac{\pi \sin \left(\frac{\pi t}{4}\right)}{4 {\cos}^{2} \left(\frac{\pi t}{4}\right)}$

$F ' \left(t\right) = \frac{\pi}{4} \tan \left(\frac{\pi t}{4}\right) \sec \left(\frac{\pi t}{4}\right)$

Integral:

$I = \int$ $\sec \left(\frac{\pi t}{4}\right)$ $\mathrm{dt}$

$I = \int$ $\sec \left(\frac{\pi t}{4}\right) \times \frac{\sec \left(\frac{\pi t}{4}\right) + \tan \left(\frac{\pi t}{4}\right)}{\sec \left(\frac{\pi t}{4}\right) + \tan \left(\frac{\pi t}{4}\right)}$ $\mathrm{dt}$

$I = \frac{4}{\pi} \int$ $\frac{\frac{\pi}{4} {\sec}^{2} \left(\frac{\pi t}{4}\right) + \frac{\pi}{4} \sec \left(\frac{\pi t}{4}\right) \tan \left(\frac{\pi t}{4}\right)}{\sec \left(\frac{\pi t}{4}\right) + \tan \left(\frac{\pi t}{4}\right)}$ $\mathrm{dt}$

$I = \frac{4}{\pi} \ln \left(\sec \left(\frac{\pi t}{4}\right) + \tan \left(\frac{\pi t}{4}\right)\right) + C$