# How do you integrate int sec^3(pix)?

May 7, 2018

$I = \frac{1}{\pi} \left[\tan \frac{\pi x}{2} \sec \left(\pi x\right) + \frac{1}{2} \ln | \tan \left(\pi x\right) + \sec \left(\pi x\right) |\right] + c$

#### Explanation:

Here,

$I = \int {\sec}^{3} \left(\pi x\right) \mathrm{dx}$

$= \int \sec \left(\pi x\right) {\sec}^{2} \left(\pi x\right) \mathrm{dx}$

$I = \int \sqrt{1 + {\tan}^{2} \left(\pi x\right)} {\sec}^{2} \left(\pi x\right) \mathrm{dx}$

Let,$\tan \left(\pi x\right) = u \implies {\sec}^{2} \left(\pi x\right) \cdot \pi \mathrm{dx} = \mathrm{du}$

$\implies {\sec}^{2} \left(\pi x\right) \mathrm{dx} = \frac{1}{\pi} \mathrm{du}$

So,

$I = \frac{1}{\pi} \int \sqrt{1 + {u}^{2}} \mathrm{du}$

$= \frac{1}{\pi} \left[\frac{u}{2} \sqrt{1 + {u}^{2}} + \frac{1}{2} \ln | u + \sqrt{1 + {u}^{2}} |\right] + c$

Subst.$, u = \tan \left(\pi x\right)$

=1/pi[tan(pix)/2sqrt(1+tan^2(pix))+1/2ln|tan(pix)+sqrt(1+tan^2(pi x))|]+c

$I = \frac{1}{\pi} \left[\tan \frac{\pi x}{2} \sec \left(\pi x\right) + \frac{1}{2} \ln | \tan \left(\pi x\right) + \sec \left(\pi x\right) |\right] + c$