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IfI=int(sin2x)/((3+4cosx)^3)dx ,t h e nI...

`IfI=int(sin2x)/((3+4cosx)^3)dx ,t h e nIe q u a l s` `(3cosx+8)/((3+4cosx)^2)+C` (b) `(3+8cosx)/(16(3+4cosx)^2)+C` `(3cosx)/((3+4cosx)^2)+C` (d) `(3-8cosx)/(16(3+4cosx)^2)+C`

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
B
`I= int (sin 2x)/((3+4cosx)^(3))dx`
Put `3+4cosx=t," so that " -4sinx dx=dt.` Then
`I=(-1)/(8)int((t-3))/(t^(3))dt=(1)/(8)((1)/(t)-(3)/(2)(1)/(t^(2)))+C`
`=(2t-3)/(16t^(2))=(8cosx+3)/(16(3+4cosx)^(2))+C`
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