# Math Help - Find the area between two trig functions

1. ## Find the area between two trig functions

Region R is enclosed by the graphs of y=tanx^2, y=(1/2)secx^2. Find R. Take it from [0,1]

For the integral I got (1/2)tanx-tanx-x.
Taking the integral from [0,1] I got about .22129. Is that right?

2. Originally Posted by ment2byours
Region R is enclosed by the graphs of y=tanx^2, y=(1/2)secx^2. Find R. Take it from [0,1]

For the integral I got (1/2)tanx-tanx-x.
Taking the integral from [0,1] I got about .22129. Is that right?
you are aware that the curves intersect at $\frac{\pi}{4}$ , right?

$A = \int_0^{\frac{\pi}{4}} \frac{\sec^2{x}}{2} - \tan^2{x} \, dx + \int_{\frac{\pi}{4}}^1 \tan^2{x} - \frac{\sec^2{x}}{2} \, dx$

$A = .3495$