Question

一道微积分问题

∫0-pai (sinx)^6*(cosx)^4 dx=?

Answer 1

(sinx)^6(cosX)^4
=(1/32) [(1-cos2x)^3(1+cos2x)^2]
=(1/32) [1-(cos2x)^2]^2(1-cos2x)
=(1/32)[(sin2x)^4-(sin2x)^2(cos2x)]

∫(sin2x)^2 dx = (1/4) ∫(1-cos4x)^2dx = (1/4) [(1-2cos4x+cos4x^2)]dx
=(1/4)[x - (1/2)sin4x + (1/2)x + (1/16)sin8x]
=(3/8)x - (1/8)sin4x + (1/64)sin8x
∫(sin2x)^2(cos2x) dx = ∫(sin 2x)^2 d(sin 2x) = (1/3) (sin 2x)^3
所以，原式：
（1/32)[ (3/8)x - (sin 4x)/8 + (sin8x)/32 - ( sin 2x)^3/3]
代入0和π：得
∫0-pai (sinx)^6*(cosx)^4 dx=　3π／(2^8) = 3π/256 = 0.0368

汗！！！真够麻烦的！

Answer 2

原式=(1/16)∫[0,π](2sinxcosx)^4*(sinx)^2dx
=(1/16)∫[0,π](sin2x)^4*(sinx)^2dx
=(1/32)∫[0,π](sin2x)^4[1-cos2x)dx
=(1/128)∫[0,π](1-cos4x)^2dx-(1/64)∫(sin2x)^2d(sin2x)
=(1/128)∫[0,π][1-2cos4x+(cos4x)^2]dx-(1/64)(sin2x)^3/3[0,π]
=(x/128)[0,π]-(1/256)sin4x[0,π]+(1/256)∫[0,π](1+cos8x)dx
=π/128-0+(x/256)[0,π]+1/(2048)sin8x[0,π]
=3π/256.

Answer 3

∫0-pai (sinxcosx)^4 sinx^2dx
=∫0-pai 1/16(2sinxcosx)^4 *1/2(1-cos2x)dx
=1/32∫0-pai sin2x^4dx-1/32∫0-pai sin2x^4*cos2xdx
=1/128∫0-pai (1-cos4x)^2dx-1/64∫0-pai sin2x^4dsin2x
=1/128∫0-pai (1+cos4x^2-2cos4x)dx-1/64*1/5sin2x^5(0-pai )
=1/128∫0-pai (1+(cos8x+1)/2-2cos4x)dx-0
=1/128(x+1/16sin8x+1/2x-1/2sin4x)(0-pai )
=3pai/256

Answer 4

先看不定积分的计算：利用
sinxcosx = (1/2)sin2x， (sinx)^2 = (1/2)(1-cos2x)，
可得
∫(sinx)^6*(cosx)^4 dx

= ∫{(1/2)(1-cos2x)[(1/2)sin2x]^4}dx
= (1/32)∫[(sin2x)^4]dx - (1/32)∫[(sin2x)^4]cos2xdx，
而
∫[(sin2x)^4]cos2xdx = (1/10) (sin2x)^5+C1，

∫[(sin2x)^4]dx = ∫[ (1/2)(1-cos4x)]^2dx = ……（降阶，直到出现奇数次幂，就可积分）
有了原函数，再利用Newton-Leibniz公式，即得所要结果。
注：在这儿写真是太痛苦了，余下的留给你了。