If `I=int(sin2x)/((3+4cosx)^(3))dx` , then I=A. `(3cosx+8)/((3+4cosx)^(2))+C`B. `(3+8cosx)/(16(3+4cosx)^(2))+C`C. `(3+cosx)/((3+4cosx)^(2))+C`D. `(3-8 cosx)/(16(3+4cosx)^(2))+C`

If `I=int(sin2x)/((3+4cosx)^(3))dx` , then I=A. `(3cosx+8)/((3+4cosx)^

1.	If `I=int(sin2x)/((3+4cosx)^(3))dx` , then I=A. `(3cosx+8)/((3+4cosx)^(2))+C`B. `(3+8cosx)/(16(3+4cosx)^(2))+C`C. `(3+cosx)/((3+4cosx)^(2))+C`D. `(3-8 cosx)/(16(3+4cosx)^(2))+C`
Answer» Correct Answer - b We have, `I=2int(cosxsinx)/((3+4cos)^(3))dx` `rArrI=-(1)/(2)int(cosx)/((3+4cosx)^3)d(3+4cosx)` `rArrI=-(1)/(2)int(t-3)/(4t^(3))dt " where " t=3+4cosx` `rArrI=(1)/(8)((1)/(t)-(3)/(2t^(2)))+C` `rArrI=(2t-3)/(16t^(2))+C=(8 cosx+3)/(16(3+4cosx)^(2))+C`

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If `I=int(sin2x)/((3+4cosx)^(3))dx` , then I=A. `(3cosx+8)/((3+4cosx)^(2))+C`B. `(3+8cosx)/(16(3+4cosx)^(2))+C`C. `(3+cosx)/((3+4cosx)^(2))+C`D. `(3-8 cosx)/(16(3+4cosx)^(2))+C`

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