Question

求sinx四次方的不定积分， 和cosx四次方的不定积分。

过程详细些，最好是方法简便

Answer 1

2 ∫(cosx)^4 dx
=∫(1-sinx^2)cosx^2dx
=∫cosx^2dx-∫sinx^2cosx^2dx
=∫(1/2)(1+cos2x)x-∫(1/4)[(1-cos4x)/2]dx
=(x/2)+(1/4)sin2x-(x/8)+(1/32)sin4x+C
=3x/8+(1/4)sin2x+(1/32)sin4x+C
1 ∫ dx/(sinx)^4
=∫ (cscx)^4 dx
consider
∫ (cscx)^4 dx=-∫ (cscx)^2 dcotx
=-(cscx)^2. cotx -2∫ (cotx)^2 (cscx)^2 dx
=-(cscx)^2. cotx -2∫ [ (cscx)^2-1] (cscx)^2 dx
5∫ (cscx)^4 dx =-(cscx)^2. cotx +2∫(cscx)^2 dx
=-(cscx)^2. cotx -2cotx
∫ (cscx)^4 dx = -(1/5) cotx [ (cscx)^2 +2 ] + C
∫ dx/(sinx)^4
=∫ (cscx)^4 dx
= -(1/5) cotx [ (cscx)^2 +2 ] + C

Answer 2

∫(sinx)^4dx = (1/4)∫[2(sinx)^2]^2dx = (1/4)∫(1-cos2x)^2]dx
= (1/4)∫[1-2cos2x+(cos2x)^2]dx = (1/4)∫[3/2-2cos2x+(1/2)cos4x]dx
= (1/4)∫[3x/2-sin2x+(1/8)sin4x] + C ;
∫(cosx)^4dx = (1/4)∫[2(cosx)^2]^2dx = (1/4)∫(1+cos2x)^2]dx
= (1/4)∫[1+2cos2x+(cos2x)^2]dx = (1/4)∫[3/2+2cos2x+(1/2)cos4x]dx
= (1/4)∫[3x/2+sin2x+(1/8)sin4x] + C

Answer 3

分享一种解法。设I1=∫(cosx)^4dx，I2=∫(sinx)^4dx。
∴I1-I2=∫[(cosx)^4-(sinx)^4]dx=∫(cos²x-sin²x)dx=∫cos2xdx=(1/2)sin2x+C1①。
I1+I2=∫[(cosx)^4+(sinx)^4]dx=∫(1-2cos²xsin²x)dx=(1/4)∫(3+cos4x)dx=3x/4+(1/16)sin4x+C2②。
联解①、②可得，I1=3x/8+(1/4)sin2x+(1/32)sin4x+C。I2=3x/8-(1/4)sin2x+(1/32)sin4x+C。
供参考。