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inte^(ax)cosbx dx =...

`inte^(ax)cosbx dx =`

A

`(e^(ax))/r cos (bx-alpha)`

B

`(e^(ax))/r cos (bx+alpha)`

C

`(e^(ax))/r sin (bx-alpha)`

D

`(e^(ax))/r sin (nx+alpha)`

Text Solution

Verified by Experts

The correct Answer is:
A
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