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If I=int(sin2x)/((3+4cosx)^(3))dx, then ...

If `I=int(sin2x)/((3+4cosx)^(3))dx`, then I equals a)`(3cosx+8)/((3+4cosx)^(2))+C` b)`(3+8cosx)/(16(3+4cosx)^(2))+C` c)`(3+cosx)/((3+4cosx)^(2))+C` d)`(3-8cosx)/(16(3+4cosx)^(2))+C`

A

`(3cosx+8)/((3+4cosx)^(2))+C`

B

`(3+8cosx)/(16(3+4cosx)^(2))+C`

C

`(3+cosx)/((3+4cosx)^(2))+C`

D

`(3-8cosx)/(16(3+4cosx)^(2))+C`

Text Solution

Verified by Experts

The correct Answer is:
B
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