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 (cos x - sinx)/(8- sin 2x) = (1)/(log) [(3sin x + cos x)/(3-sin x - cos x)] + c`
Now `int(cos x -sinx)/(8-sin 2x)dx`
` int = (cos x - sinx)/(9-(1+(2 sin x cos x)) dx`
`int(cos x-sinx)/((3)^(2) -(cos x+sinx)^(2))dx`
`=int(cos x - sinx)/((3)^(2) -(cos x + sinx)^(2))dx`
put cos x + sin t
`(-sin x+ cosx) dx = dt`
` = int(dt)/((3)^(2) -(t)^(2)) = (1)/(2(3)) log|(3+t)/(3-t)|+C`
` = (1)/(6) log|(3+sin x+cosx)/(3-sin - cos x)| +C`
`therefore` p = 6