`int (cos 5x + cos 4x)/(1-2cos3x) dx`

Answer

N/a

Explanation

`I = int(cos5x+cos4x)/(1-2cos3x)dx=int(2cos'(9x)/(2).cos'x/2)/(1-2(2cos^(2)'(3x)/(2)-1))dx`
`[:' cosC + cosD = 2cos'(C-D)/(2).cos'(C-D)/(2) "and" cos2x = 2cos^(2)x-1]`
`:. I= -int(2cos'(9x)/2.cos'x/2)/(3-4cos^(2)'(3x)/(2))dx=-int(2cos'(9x)/(2).cos'(x)/(2))/(4cos^(2)'(3x)/(2)-3)dx`
`=-int(2cos'(9x)/(2).cos'(x)/(2).cos'(3x)/(2))/(4cos^(3)'(3x)/(2)-3cos'(3x)/(2)).dx` [multiply and divide by `cos= (3x)/(2)`]
`=-int(2cos'(9x)/(2).cos'(3x)/(2))/(cos3.'(3x)/(2))dx = -int2cos'(3x)/(2).cos'x/2dx`
` =-int{cos((3x)/(2)+
x/2)+cos((3x)/(2)-x/2)}dx`
`-int(cos2x+cosx)dx`
`=-[(sin2x)/(2)+sinx]+C`
`=-1/2 sin2x- sinx + C`