Why Is My Integral Result Incorrect for Cosine of 2 Theta?

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The integral of cosine of 2 theta from pi minus theta1 to pi plus theta1 was initially evaluated incorrectly, resulting in a positive one half sine of 2 theta1 instead of the expected negative value. The error was traced back to the orientation of the vectors used in the calculation. By reversing the direction of the vector, the interval also changes, leading to the correct result. The discussion highlights the importance of vector orientation in integral evaluations. Understanding these concepts is crucial for accurate calculations in trigonometric integrals.
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Homework Statement



From angles pi minus theta1 to pi plus theta1, what is the integral of cosine of 2 theta times d theta?

Homework Equations





The Attempt at a Solution



When I evaluate this definite integral from pi minus theta1 to pi plus theta1, I get positive one half sine of 2 theta1.
It should be NEGATIVE one half sine of 2 theta1.

Is 2pi + 2theta1 and 2pi - 2theta1 somehow wrong?
 
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Hi PieOperator! :smile:

(have a pi: π and a theta: θ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)

you should have got [1/2 sin2θ]π-θ1π+θ1 …

what happened then? :confused:​
 
Thanks. I found my mistake. It has to do with vectors and the correct orientation. It is positive one half sine of 2 theta in the incorrect vector orientation. I just have to reverse the direction of the vector, and the interval reverses as well.
 

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