∫ cos⁶x dx

= (1/8)∫ (1 + 3cos2x + 3cos²2x + cos³2x) dx

= (1/8)∫ dx + (3/8)∫ cos2x dx + (3/8)∫ cos²2x dx + (1/8)∫ cos²2x cos2x dx

= x/8 + (3/8)(1/2)sin2x + (3/8)(1/2)∫ (1 + cos4x) dx + (1/8)(1/2)∫ cos²2x dsin2x

= x/8 + (3/16)sin2x + (3/16)(x + 1/4 · sin4x) + (1/16)∫ (1 - sin²2x) dsin2x

= x/8 + (3/16)sin2x + 3x/16 + (3/64)sin4x + (1/16)[sin2x - (sin³2x)/3] + C

= 5x/16 + (1/4)sin2x + (3/64)sin4x - (1/48)sin³2x + C

即求(cost)^5的不定積分.(cost)^5dt

=(cost)^4dsint

=[(cost)^2 ]^2dsint

=[1-(sint)^2]^2dsint

=[1-2sin^2t+(sint)^4]dsint，

可計算其原函式為

sint-2/3(sint)^3+1/5(sint)^5+C