Calculus Definite Integrals of Trig Functions
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The Sinister Trig Integral
Definite Integrals of Trig Functions


 What if we wanted to find the area between the X axis and the function ...


 F(x) = sinX + cosX

 and we want all of the area from X = 0 to X = p radians.
 (for those not fluent in radian-ese, that's from 0 to 180 degrees)
 The first thing we need to do is figure out if the function is always positive,
 or always negative, or crosses the axis, or what.
 When X = 0, cosX = 1 and sinX = 0. 
 So the total function value is 1.
 That means our function starts out ABOVE the X axis
 Now lets see if the function ever touches the X axis?
 That is, is there a point where:

 sinX + cosX = 0

 Let's find out ...

 sinX + cosX = 0

 sinX = - cosX

 In quadrant II (90 to 180), sine is positive and cosine is negative.
 The sine and cosine are the same (except maybe for sign on the number) 
 at 45, 135, 215, 305 (45 + 90 x n).
 The version of that in quadrant II is 135, also known as 3p/4 radians.
 So we have two sections. One from 0 to 3p/4 radians 
 and one from 3p/4 radians  to p radians.
 Since the graph line is above the X axis at X = 0,
 we know the first section is ok. That is, that area will turn out positive.
 How do we tell if the second section will be positive or negative?
 Take any point in that section (excluding 3p/4 ), 
 and test to see if the function has a positive or negative value.
 So let's test the function at p.

 F(x) = sinX + cosX

 F(p) = sinp + cosp

F(p) = 0 - 1

 F(p) = - 1

 So the section of the curve to the right of 3p/4 is below the X axis.
 The graph looks something like this:


 So the problem works out this way ...


 That didn't hurt all THAT much did it?
 Well at least it's over ...

   copyright 2005 Bruce Kirkpatrick

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