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int(sin2x)/(a+bcosx)^2dx...

`int(sin2x)/(a+bcosx)^2dx`

Text Solution

Verified by Experts

The correct Answer is:
` -(2)/(b^(2))[log|a+b cosx|+(a)/(a+b cosx)]+C`
Let ` I=int(sin2x)/(a+bcosx)^(2)dx=int(2sinx cosx)/(a+bcosx)^(2)dx.`
` " Putting " a+bcosx=t " and "-bsinxdx=dt, " we get " `
` I=-(2)/(b)int(1)/(t^(2))((t-a)/(b))dt [ :' a+bcosx=t, cosx=(t-a)/(b)] `
`=-(2)/(b^(2))int((1)/(t)-(a)/(t^(2)))dt=-(2)/(b^(2))[log|t|+(a)/(t)]+C`
`= -(2)/(b^(2))[log|a+b cosx|+(a)/(a+b cosx)]+C`
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Knowledge Check

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