- #1

PrudensOptimus

- 641

- 0

sin3x, with respect to x. [-a,a] - interval

I end up getting 0/3:

= [-cos(3x)/3]

= [-cos(3a) - (- cos (-3a))]/3

= [-cos (3a) + cos (3a)]/3 ---> cos(-3a) = cos 3a

= 0/3 = 0

I think I did something wrong, right?

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- Thread starter PrudensOptimus
- Start date

- #1

PrudensOptimus

- 641

- 0

sin3x, with respect to x. [-a,a] - interval

I end up getting 0/3:

= [-cos(3x)/3]

= [-cos(3a) - (- cos (-3a))]/3

= [-cos (3a) + cos (3a)]/3 ---> cos(-3a) = cos 3a

= 0/3 = 0

I think I did something wrong, right?

- #2

Suicidal

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