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Integration Question (1 Viewer)

tigerdedman

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mmm i'm not entirely sure but i'm thinking u would just square it ie. square sin x to negate the postive and negative case.

this would give u the standard 3 unit integration of (sin x)squared, whereby u use the cos double angle formula to integrate it. then once u have your answer just sqareroot the answer, so as ur not actually changing the question.
 

Mattamz

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is this as a definite integral?

is so you would just add together the integral above the x-axis (say between 0 and pi) and the absolute value of the integral below the x-axis (say between pi and 2pi)
 

onebytwo

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tigerdedman said:
mmm i'm not entirely sure but i'm thinking u would just square it ie. square sin x to negate the postive and negative case.

this would give u the standard 3 unit integration of (sin x)squared, whereby u use the cos double angle formula to integrate it. then once u have your answer just sqareroot the answer, so as ur not actually changing the question.
i dont think that approach is mathematically acceptable. consider the following:

e.g. int. (x) dx = (x^2)/2 +c

now is this the same as sqrt. [int. (x^2) dx]

no, because that yields sqrt[(x^3)/3] +c1 =/= (x^2)/2 +c2.

i think what needs to be done is the function must be broken down into two cases: sin (x) for 0<=x<=pi and -sin (x) for pi<x<2pi. then the integral found for both cases.

remember that : abs. (x) = x, for x>=0 and -x for x<0
 

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