Question

What is the integral of `sqrt(cos^(2)x)` from zero to pi?

Answer 1

`int_0^ pi sqrt cos^2 xdx`= `int_0^pi cos x dx`

This definite integral means the area bounded by the curve y= cos x between x=0 to x= `pi`. To evaluate this area correctly, divide the interval in two parts 0 to `pi/2` and `pi/2` to `pi`
= `[sinx]_0^(pi/2)` +

`[sin x]_(pi/2)^(pi)`

The area above x axis would come to be +1 and that below x axis would come to be -1. The sum of the two areas would not be 0.
The sum of both the areas would be 1+1 =2

Answer 2

Note that `sqrt(u^2) = absu` if we cannot be sure that `u` is positive.

In the interval `[0, pi]`, we have:.

`sqrt(cos^2x ) = abs(cosx) = { (cosx, ", if " 0<= x <= pi/2), (-cosx, ", if " pi/2<= x <= pi) :}`

So the integral becomes:

`int_0^pi sqrt(cos^2x ) dx = int_0^pi abs(cosx) dx`

`color(white)"sssssssssssssss"` `= int_0^(pi/2) cosx dx+int_(pi/2)^pi -cosx dx`

Both of these integrals evaluate to `1`, so we get:

`int_0^pi sqrt(cos^2x ) dx = 1+1=2`