[（sin x-cos x）/(sin x+cos x)]^4 求定积分 积分区间0-π/4

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[（sin x-cos x）/(sin x+cos x)]^4 求定积分 积分区间0-π/4

[（sin x-cos x）/(sin x+cos x)]^4 求定积分 积分区间0-π/4
[（sin x-cos x）/(sin x+cos x)]^4 求定积分 积分区间0-π/4

[（sin x-cos x）/(sin x+cos x)]^4 求定积分 积分区间0-π/4
[（sin x-cos x）/(sin x+cos x)]^4 =[(1-tgx)/(1+tgx)]^4=[tg(x-45)}^4=[sec^2(x-45)-1]^2
由此再求,上面两答案都不对

∫[（sin x-cos x）/(sin x+cos x)]^4dx
=-∫(sin x+cos x)]^(-4)d(sin x+cos x)
=1/3*(sin x+cos x)^(-3)+c
积分区间0-π/4
=1/3*[(sin π/4+cosπ/4)^(-3)-(sin 0+cos0)^(-3)]
=1/3*(√2/4-1)
希望能帮到你...

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∫[（sin x-cos x）/(sin x+cos x)]^4dx
=-∫(sin x+cos x)]^(-4)d(sin x+cos x)
=1/3*(sin x+cos x)^(-3)+c
积分区间0-π/4
=1/3*[(sin π/4+cosπ/4)^(-3)-(sin 0+cos0)^(-3)]
=1/3*(√2/4-1)
希望能帮到你，祝学习进步O(∩_∩)O,也别忘了采纳！

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∫[0→π/4] [(sin x-cos x)/(sin x+cos x)]⁴dx
= ∫[0→π/4] d(-cos x-sin x)[1/(sin x+cos x)]⁴dx
= -∫[0→π/4] d(cos x+sin x)[1/(sin x+cos x)]⁴dx
= (1/3)∫[0→π/4] d[1/(sin x+cos x)]...

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∫[0→π/4] [(sin x-cos x)/(sin x+cos x)]⁴dx
= ∫[0→π/4] d(-cos x-sin x)[1/(sin x+cos x)]⁴dx
= -∫[0→π/4] d(cos x+sin x)[1/(sin x+cos x)]⁴dx
= (1/3)∫[0→π/4] d[1/(sin x+cos x)]³dx
= (1/3)[1/(sin x+cos x)]³ [0→π/4]
= [（根号2）/4 - 1]/3
= (根号2)/12 - 1/3

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