If `int(cosx-sinx)/(8-sin2x)dx=(1)/(p)log[(3+sinx+cosx)/(3-sinx-cosx)]+c`, then p= . . . .

Options

A. 6
B. 1
C. 3
D. 12

Answer

A

Explanation

We have, `int(cosx-sinx)/(8-sin2x)=(1)/(p)log[(3+sinx+cosx)/(3-sinx-cosx)]+C`
Now, `int(cosx-sinx)/(8-sin2x)dx`
`=int(cosx-sinx)/(9-(1+2sinxcosx))dx`
`=int(cosx-sinx)/(9-(sin^(2)x+cos^(2)x+2sinxcosx))dx`
`=int(cosx-sinx)/((3)^(2)-(cosx+sinx)^(2))dx`
Put cosx+sinx=t
`(-sinx+cosx)dx=dt`
`=int(dt)/((3)^(2)-(t)^(2))=(1)/(2(3))"log"|(3+t)/(3-t)|+C`
`=(1)/(6)"log"|(3+sinx+cosx)/(3-sinx-cosx)|+C`
`thereforep=6`