The given function is cos(2x) ( cosx+sinx ) 2 .
The given function can be written as,
cos(2x) ( cosx+sinx ) 2 = cos(2x) cos 2 x+ sin 2 x+2sinxcosx = cos(2x) 1+sin2x (1)
From (1), we get,
∫ cos(2x) ( cosx+sinx ) 2 dx = ∫ cos( 2x ) 1+sin(2x) dx
Put 1+sin( 2x )=t 2cos( 2x )dx=dt
Substitute t for 1+sin(2x) and dt for 2cos( 2x )dx in (1),
∫ cos(2x) ( cosx+sinx ) 2 dx = 1 2 ∫ 1 t dt = 1 2 log| t |+c = 1 2 log| 1+sin(2x) |+c = 1 2 log| ( sin(x)+cos(x) ) 2 |+c =log( sinx+cosx )+c
Thus, the integral of the function cos(2x) ( cosx+sinx ) 2 is log( sinx+cosx )+c.