integration tutorial: analysis

Consider the integral
int 02p sin(t)2 dt
This is a definite integral, so we need to find the area under the integrand f(t), which we do by using the Fundamental Theorem of Calculus: we find an antiderivative, evaluate it at the endpoints of the integral, and take the difference of the values.
To find an antiderivative of f(t), we go through our list of integration methods:
  1. Recognize elementary antiderivatives
  2. Rewrite the integrand to make it easier
  3. Use substitution to reverse the chain rule or simplify the integrand
  4. Use integration by parts
  5. Use inspection to see the value of a definite integral
to find one that works for this integrand. In this case, only one method is appropriate. We can rewrite the integrand: use the half-angle formula to give an expression that is more easily integrable. This is shown below.
Rewriting:
Use the half-angle formula to rewrite the integral:
sin(t)2 = (1-cos(2t))/2.
This can be broken into separate integrals,
int sin(t)2 dt =int1/2dt+int-cos(2t)/2dt
which can be evaluated using basic antidifferentiation, and substitution (of "ax"), respectively, to obtain
int1/2dt+int-cos(2t)/2dt = (t/2) + (-(1/4)sin(2t)) + C.

Explanation for rewritten term(s)
Basic antidifferentiation
In this case, by using rules for basic antiderivatives we can just write down the antiderivative of the integrand:
int 1/2 dt = t/2 + C

Substitution (of "ax")
Let w = 2 t. Then w' = 2, so (1/2) dw = dt. The integral can therefore be rewritten as

int -cos(2t)/2 dt = int -cos(w)/2 (1/2) dw = -(1/4)sin(w) + C
Thus, substituting back for w,
int -cos(2t)/2 dt = -(1/4)sin(2t) + C

To evaluate the definite integral, we take this antiderivative, evaluate it at the endpoints of the integral (0 and 2pi), and take the difference of the values. This gives
[ ((2pi)/2) + (-(1/4)sin(2(2pi))) ] - [ ((0)/2) + (-(1/4)sin(2(0))) ] = p.
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integration analysis
Last Modified: Wed Feb 6 13:53:59 EST 2002
Comments to glarose@umich.edu
©2002 Gavin LaRose, UM Math Dept.