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Worked Out Answer 3.4 1e from: Maths in Motion – Theo de Haan.

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Presentation on theme: "Worked Out Answer 3.4 1e from: Maths in Motion – Theo de Haan."— Presentation transcript:

1 Worked Out Answer 3.4 1e from: Maths in Motion – Theo de Haan

2 For the sake of simplicity, leave out the limits of integration. Rewrite as:

3 Identify sin  (except for a minus sign) as the derivative of cos . You can also identify cos  as the derivative of sin , but unfortunately this doesn’t work out quite the same way...

4 Rewrite as: Substitute: So d  becomes: Now substitute u and d 

5 = = cos  d  = Please note that u (and therefore cos  ) has to be greater than zero. =

6 = = cos  d  = = By changing over to  again, you can use the original limits of integration:

7 = You have found: The definite integral now becomes: = =0.62 In radians!ln(1) = 0

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