Question

这定积分题的解题思路是什么

求解答

Answer 1

思路:
1、降次
2、积化和差
3、最后积分
看过程领会。

满意，请及时采纳。谢谢！

Answer 2

(sin3x)^2(sin2x)^3
=(1-cos6x)(1-cos4x)sin2x/4
=(1-cos6x-cos4x+cos6xcos4x)sin2x/4
=[2sin2x-2sin2xcos6x-2sin2xcos4x+sin2x(cos10x+cos2x)]/8
=[2sin2x-sin8x+sin4x-sin6x+sin2x+(1/2)(sin12x-sin8x+sin4x)]/8
=(6sin2x+3sin4x-2sin6x-3sin8x+sin12x)/16,
所以原式=（1/16）[-3cos2x-(3/4)cos4x+(1/3)cos6x+(3/8)cos8x-(1/12)cos12x]+c.,
还可以设u=cos2x,du=-2sin2xdx,cos6x=4u^3-3u,
原式=(-1/4)∫[1+3u-4u^3)(1-u^2)du
=(-1/4)∫(1+3u-u^2-7u^3+4u^5)du
=(-1/4)[u+(3/2)u^2-(1/3)u^3-(7/4)u^4+(2/3)u^6]+c
=(-1/4)[cos2x+(3/2)(cos2x)^2-(1/3)(cos2x)^3-(7/4)(cos2x)^4+(2/3)(cos2x)^6]+c.

Answer 3

cos(A-B)= cosA.cosB + sinA.sinB

cos(A+B)= cosA.cosB - sinA.sinB

=>

sinA.sinB = (1/2)[ cos(A-B)-cos(A+B) ]

A=3x, B=2x

sin3x. sin2x = (1/2)[ cosx -cos5x]

(sin3x.sin2x)^2 

=(1/4)[ cosx -cos5x]^2

=(1/4) [( cosx)^2-2cosx.cos5x + (cos5x)^2 ]

=(1/4) [ (1/2)(1+ cos2x)-[cos4x+cos6x] + (1/2)(1+cos10x) ]

=(1/8) [ 2 + cos2x -cos4x-cos6x +cos10x]

(sin3x.sin2x)^2 . sin2x

=(1/8) [ 2 + cos2x -cos4x-cos6x +cos10x] . sin2x

=(1/8) [ 2sin2x + cos2x.sin2x -cos4x.sin2x-cos6x.sin2x +cos10x. sin2x]

=(1/8)[2sin2x + (1/2)sin4x -(1/2)(sin6x-sin2x) -(1/2)(sin8x-sin4x) +(1/2)(sin12x-sin8x) ] 

=(1/8)[3sin2x + sin4x -(1/2)sin6x -sin8x +(1/2)sin12x ] 

=(1/16)[6sin2x + 2sin4x -sin6x -2sin8x +sin12x ] 

∫(sin3x)^2 . (sin2x)^3 dx

=∫ [(sin3x).(sin2x)]^2 . sin2x dx

=(1/16)∫[6sin2x + 2sin4x -sin6x -2sin8x +sin12x ]  dx

=(1/16)[-3cos2x - (1/2)cos4x +(1/6)cos6x +(1/4)cos8x -(1/2)cos12x ]   +C