Integral sqrt(1-(cos(x))^2) on 0 to 2pi, reduces to integral abs(sin(x)) - split the integral!
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Integral sqrt(1-(cos(x))^2) on 0 to 2pi, reduces to integral abs(sin(x)) - split the integral!
In this problem, we have a definite integral sqrt(1-(cos(x))^2) on the interval 0 to 2pi. Clearly, the first move is to use the identity (sin(x))^2=1-(cos(x))^2 to simplify the interior of the square root, and this gives us sqrt((sin(x))^2) in the integral.
Can we say the square root undoes the square? We proceed naively with this idea, but the integral comes out to zero. This is impossible because the integrand is non-negative (and usually positive).
What went wrong? The problem is that the square root of a square is the absolute value, and we can't ignore that absolute value on an integration interval which includes negative values of the sine function.
To correct the problem, we split the integration interval at x=pi, where the sine function transitions from positive to negative values. On the positive interval, the absolute value does nothing, and on the negative interval, the absolute value multiplies the sine function by -1 in order to get a positive value. Now we can perform both integrals and obtain the result of 4 for the total area under the curve.
Finally, we verify with a graph of the original integrand that an area of 4 looks right.
