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求不定积分:∫ (cosx)^6 dx
人气:207 ℃ 时间:2020-02-06 08:57:33
解答
∫ (cosx)^6 dx
=∫ (cosx)^5 dsinx
= sinx(cosx)^5 + ∫ 5(cosx)^4 (sinx)^2 dx
= sinx(cosx)^5 + ∫ 5(cosx)^4 (1-(cosx)^2) dx
6∫ (cosx)^6 dx = sinx(cosx)^5 + 5∫ (cosx)^4dx
=sinx(cosx)^5 + (5/4) (4∫ (cosx)^4dx)
=sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +3∫ (cosx)^2dx )
=sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +(3/2)(2∫ (cosx)^2dx ) )
=sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +(3/2)[sinxcosx+∫ dx] )
=sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +(3/2)[sinxcosx+x] )
∫ (cosx)^6 dx = (1/6){ sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +(3/2)[sinxcosx+x] ) } + C
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