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求不定积分∫1/cos^5xdx
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求不定积分∫1/cos^5xdx
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答案和解析
I = ∫ dx/cos^5x = ∫ cosxdx/cos^6x = ∫ dsinx/(1-sin^2x)^3 = ∫ du/(1-u^2)^3
= (1/14) ∫ [3/(1-u)+3/(1+u)+3/(1-u)^2+3/(1+u)^2+1/(1-u)^3+1/(1+u)^3]du
= (3/14)ln|(1+u)/(1-u)| +(3/14)[1/(1-u)-1/(1+u)]+(1/28)[1/(1-u)^2-1/(1+u)^2]+C
= (3/7)ln|(1+sinx)/cosx| +(3/7)sinx/(cosx)^2+(1/7)sinx/(cosx)^4+C
= (1/14) ∫ [3/(1-u)+3/(1+u)+3/(1-u)^2+3/(1+u)^2+1/(1-u)^3+1/(1+u)^3]du
= (3/14)ln|(1+u)/(1-u)| +(3/14)[1/(1-u)-1/(1+u)]+(1/28)[1/(1-u)^2-1/(1+u)^2]+C
= (3/7)ln|(1+sinx)/cosx| +(3/7)sinx/(cosx)^2+(1/7)sinx/(cosx)^4+C
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