- #1

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[tex]\int[/tex] e[tex]^{i2t}[/tex] dt between t and 0.

I've tried it with trigonometric identities and keep getting lost!

Cheers!

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- Thread starter Master J
- Start date

- #1

- 226

- 0

[tex]\int[/tex] e[tex]^{i2t}[/tex] dt between t and 0.

I've tried it with trigonometric identities and keep getting lost!

Cheers!

- #2

Char. Limit

Gold Member

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Try a u-substitution.

u=2it

du=2i dt

u=2it

du=2i dt

- #3

AlephZero

Science Advisor

Homework Helper

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It is exactly the same as

[tex]\int_0^t e^{at}\,dt[/tex]

except that [tex]a[/tex] is a complex number.

- #4

Char. Limit

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- #5

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Cheers guys.

Yea true I guess, perhaps it should have been tau as the differential.

Yea true I guess, perhaps it should have been tau as the differential.

- #6

- 226

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The answer is 2(1 - coswt) / w ....i cant seem to get this at all. Any ideas?

- #7

Char. Limit

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Is it the integral that's squared, or just the integrand?

- #8

HallsofIvy

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[tex]\int_0^t e^{-2i\tau}d\tau[/tex]

is

[tex]-\fra{1}{2i}e^{-2it}= \frac{i}{2}e^{-2it}[/tex]

NOT

[tex]\frac{2(1- cos(\omega t))}{\omega}[/tex]

or even

[tex]\frac{2(1- cos(2t))}{2}= 1- cos(2t)[/itex]

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