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If I= int (sin 2x)/((3+4cosx)^(3))dx, t...

If `I= int (sin 2x)/((3+4cosx)^(3))dx,` then `I` equals

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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Knowledge Check

  • 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`

  • int(sin3x)/(cosx)dx=

    A
    `log|cosx|+2cos^2x+c`
    B
    `log|secx|+2cos^2x+c`
    C
    `log|cosx|-2cos^2x+c`
    D
    `log|secx|-2cos^2x+c`

  • int(3+2cosx)/((2+3cosx)^(2))dx is equal to

    A
    `((sinx)/(3cosx+2))+c`
    B
    `((2cosx)/(3sinx+2))+c`
    C
    `((2cosx)/(3cosx+2))+c`
    D
    `((2sinx)/(3sinx+2))+c`

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    I=inte^(cosx-sinx)(sin x+cos x)dx

    I=int (sinx+cosx)/sqrt(1-sin2x) dx

    int (sin x dx)/(3+4 cos^2 x) is equal to

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