# What is the definite integral of sec^4 x from 0 to pi/4?

Sep 30, 2014

${\int}_{0}^{\frac{\pi}{4}} {\sec}^{4} \left(x\right) \mathrm{dx} = {\int}_{0}^{\frac{\pi}{4}} {\sec}^{2} \left(x\right) {\sec}^{2} \left(x\right) \mathrm{dx}$

Trig Identity

${\sec}^{2} \left(x\right) = {\tan}^{2} \left(x\right) + 1$

Use this identity to substitute for one of the ${\sec}^{2} \left(x\right)$.

${\int}_{0}^{\frac{\pi}{4}} \left[{\tan}^{2} \left(x\right) + 1\right] {\sec}^{2} \left(x\right) \mathrm{dx}$

Now begin with u-substitution

Let $u = \tan \left(x\right)$

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

$\int \left[{u}^{2} + 1\right] \mathrm{du}$

$\left[{u}^{3} / 3 + u\right]$ Convert back to terms of x $\to {\left[\tan {\left(x\right)}^{3} / 3 + \tan \left(x\right)\right]}_{0}^{\frac{\pi}{4}}$

$= \left[\tan {\left(\frac{\pi}{4}\right)}^{3} / 3 + \tan \left(\frac{\pi}{4}\right) - \left(\tan \frac{0}{3} + \tan \left(0\right)\right)\right]$

$= \left[{\left(1\right)}^{3} / 3 + 1 - \left(0 + 0\right)\right]$

$= \left[\frac{1}{3} + 1\right]$

$= \left[\frac{1}{3} + \frac{3}{3}\right]$

$= \left[\frac{4}{3}\right]$

$= 1.3333$

Remember that, $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

So we can also say, ${\sec}^{4} \left(x\right) = \frac{1}{{\cos}^{4} \left(x\right)}$ After graphing press 2nd and then TRACE

Press 7 for integration, $\int f \left(x\right) \mathrm{dx}$

Then enter the LOWER and UPPER LIMITS

See the results of those actions below. 