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The reduction formula for ∫cosnx dx is: ∫cosnx dx = (1/n)cosn-1x sin x + ((n-1)/n)∫cosn-2x dx.First, apply the formula for n=4:∫cos4x dx = (1/4)cos3x sin x + ( (4-1)/4 )∫cos4-2x dx∫cos4x dx = (1/4)cos3x sin x + (3/4)∫cos2x dxNext, apply the formula for n=2 to evaluate ∫cos2x dx:∫cos2x dx = (1/2)cos2-1x sin x + ( (2-1)/2 )∫cos2-2x dx∫cos2x dx = (1/2)cos x sin x + (1/2)∫cos0x dx∫cos2x dx = (1/2)cos x sin x + (1/2)∫1 dx∫cos2x dx = (1/2)cos x sin x + (1/2)x + C1Now, substitute this result back into the expression for ∫cos4x dx:∫cos4x dx = (1/4)cos3x sin x + (3/4)[(1/2)cos x sin x + (1/2)x] + C∫cos4x dx = (1/4)cos3x sin x + (3/8)cos x sin x + (3/8)x + CThis result exactly matches option A.