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int(2sinx)/(3+sin2x)\ dx...

`int(2sinx)/(3+sin2x)\ dx`

A

`(1)/(2) "In"|(2+sinx-cosx)/(2-sinx+cosx)|-(1)/(sqrt(2))tan^(-1)((sinx+cosx)/(sqrt(2)))+c`

B

`(1)/(2) "In"|(2+sinx-cosx)/(2-sinx+cosx)|-(1)/(2sqrt(2))tan^(-1)((sinx+cosx)/(sqrt(2)))+c`

C

`(1)/(4) "In"|(2+sinx-cosx)/(2-sinx+cosx)|-(1)/(sqrt(2))tan^(-1)((sinx+cosx)/(sqrt(2)))+c`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`I=int(2sinx)/((3+sin2x))dx`
`=int(sinx+cosx+sinx-cosx)/((3+sin2x))dx`
`=int (sinx+cosx)/(3+underset(I_(1))(underset(darr)(sin)) 2x)dx-int (-sinx+cosx)/((3+underset(I_(2))(underset(darr)(sin))2x))dx`
Putting `t_(1)=sinx-cosx " in "I_(1) " and " I_(2)=sinx+cosx " in " I_(2),` we get
`I=int(dt_(1))/([3+(1-t_(1)^(2))])-int (dt_(2))/([3+(t_(2)^(2)-1)])=int (dt_(1))/(4-t_(1)^(2))-int(dt_(2))/(2+t_(2)^(2))`
`=(1)/(4)"In"|(2+t_(1))/(2-t_(1))|-(1)/(sqrt(2))tan^(-1)((t_(2))/(sqrt(2)))+C`
`=(1)/(4) "In"|(2+sinx-cosx)/(2-sinx+cosx)|-(1)/(sqrt(2))tan^(-1)((sinx+cosx)/(sqrt(2)))+C`
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