∫sin(8次方)xdx

Question

∫sin(8次方)xdx

Answer 1

sin^8(x)
= (sin²x)⁴
= [(1-cos2x)/2]⁴
= (1/16)(1-4cos2x+6cos²2x-4cos³2x+cos⁴2x)
= (1/16) - (1/4)cos2x + (3/8)cos²2x - (1/4)cos³2x + (1/16)cos⁴2x
= (1/16) - (1/4)cos2x + (3/16)(cos4x+1) - (1/8)(cos4x+1)cos2x + (1/64)(cos4x+1)²
= (1/4) - (3/8)cos2x + (3/16)cos4x - (1/8)cos4xcos2x + (1/64)(cos²4x+2cos4x+1)
= (17/64) - (3/8)cos2x + (7/32)cos4x - (1/16)(cos6x+cos2x) + (1/64)cos²4x
= (35/128) - (7/16)cos2x + (7/32)cos4x - (1/16)cos6x + (1/128)cos8x
∫sin^8(x) dx
= ∫[(35/128) - (7/16)cos2x + (7/32)cos4x - (1/16)cos6x + (1/128)cos8x] dx
= (35/128)x - (7/32)sin2x + (7/128)sin4x - (1/96)sin6x + (1/1024)sin8x + C
答案已经经过软件验算,绝对正确.