Area between trigo function and a line How can I compute the area of region enclosed by $\cos^2x$ and $y=1-\frac{2}{\pi}x$?
I am just stuck on the first step, finding their intersection point, I know I can plot the graph using online resources but is there any numerical method to obtain the intersection?
 A: Define $f(x)=\cos^2x$ and $g(x)=1-2/\pi \cdot x$.

Two intersection points can simply be found by plugging in $x=0$ and $x=\pi/2$. For the third intersection point, you may use numerical methods such Newton-Raphson, Secant and the alike. But if this is a problem on a test which it seems to be because of the "nice" $x$-values we're getting as intersection points. You might try out other special angles and you'll observe that $x=\pi/4$ is an intersection point as well.
All you're now left to do is to figure out in what intervals, what function is greater and integrate from the suitable points.

Clearly, $g(x)$ has a negative slope for all $x$ and the derivative of $\cos^2x$ is $-\sin2x$ which tells us it is zero at $x=0$ and $x=\pi/2$. Plug in any value between $0$ and $\pi/2$ to figure out the sign of the slope of $f(x)$ say $x=\pi/4$, which tells you this is negative too.
Plug in any values between $x=0$ and $x=\pi/4$ to figure out which function would be greater in $[0, \pi/4]$ and do the same for the interval $[0,\pi/2]$.

$$\text{Area}=\int_{0}^{\pi/4}\left(\cos^2x-1+\dfrac{2}{\pi}x\right)\mathrm dx+\int_{\pi/4}^{\pi/2}\left(1-\dfrac{2}{\pi}x-\cos^2x\right)\mathrm dx $$
