- #1

blackbear

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1. T[Cos(t)] = ∫[Sin(t-x)Cos(t)]dx the limits are from 0 to 2pi

2. T[Sin(t)]=∫[Sin(t-x)Sin(t)]dx; the limits are from 0 to 2pi

Thanks

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- Thread starter blackbear
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- #1

blackbear

- 41

- 0

1. T[Cos(t)] = ∫[Sin(t-x)Cos(t)]dx the limits are from 0 to 2pi

2. T[Sin(t)]=∫[Sin(t-x)Sin(t)]dx; the limits are from 0 to 2pi

Thanks

- #2

Mark44

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What have you tried? You need to make a reasonable effort before we can provide any help.

- #3

berkeman

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1. T[Cos(t)] = ∫[Sin(t-x)Cos(t)]dx the limits are from 0 to 2pi

2. T[Sin(t)]=∫[Sin(t-x)Sin(t)]dx; the limits are from 0 to 2pi

Thanks

Per the PF rules, you need to show some effort at solving these problems before we can help you. But I'll offer a hint: If the integration is with respect to x (as indicated by the dx in each equation), then what can you do with the Sin(t) and Cos(t) terms in each equation?

- #4

blackbear

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Well Sin(t) and Cos(t) is a constant; each of these terms will come outside the integration, if we x is the integrating variable.

Last edited:

- #5

blackbear

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I computed both the integration and getting "0" for both results!!!

- #6

berkeman

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I computed both the integration and getting "0" for both results!!!

Probably related to integrating sinusoidal functions from 0 to 2*PI, eh?

- #7

blackbear

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yes...the integrating result with 0 to pi gave me the "zero" results.

- #8

sEsposito

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You really should post your attempted solution so that we can see what's plaguing you...

- #9

berkeman

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yes...the integrating result with 0 to pi gave me the "zero" results.

No, you said 0 to 2*PI in your original post (OP) above.

- #10

blackbear

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regards

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