这个积分怎么积呢??
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令cosx=u,则:∫√[1+(cosx)^2]d(cosx)=∫√(1+u^2)du。
再令u=tant,则:du=[1/(cost)^2]dt。
∴∫√[1+(cosx)^2]d(cosx)
=∫√[1+(tant)^2][1/(cost)^2]dt
=∫[1/(cost)^3]dt
=∫[cost/(cost)^4]dt
=∫[1/(cost)^4]d(sint)
=∫{1/[1-(sint)^2]^2}d(sint)。
令sint=y,则:
∫√[1+(cosx)^2]d(cosx)
=∫[1/(1-y^2)^2]dy
=(1/4)∫[1/(1-y)+1/(1+y)]^2dy
=(1/4)∫[1/(1-y)^2]dy+(1/2)∫[1/(1-y^2)]dy+(1/4)∫[1/(1+y)^2]dy
=(1/4)/(1-y)-(1/4)/(1+y)+(1/4)∫[1/(1-y)+1/(1+y)]dy
=(1/4)[(1+y)-(1-y)]/(1-y^2)-(1/4)ln|1-y|+ln|1+y|+C
=(1/2)y/(1-y^2)+(1/4)ln|(1+y)/(1-y)|+C
=(1/2)sint/[1-(sint)^2]+(1/4)|(1+sint)/(1-sint)|+C
=(1/2)tant/cost+(1/4)ln|(1+sint)^2/(cost)^2|+C
=(1/2)u√[1+(tant)^2]+(1/2)ln[(1/cost+tant]+C
=(1/2)u√(1+u^2)+(1/2)ln{√[1+(tant)^2]+u}+C
=(1/2)cosx√[1+(cosx)^2]+(1/2)ln{√[1+(cosx)^2]+cosx}+C。
再令u=tant,则:du=[1/(cost)^2]dt。
∴∫√[1+(cosx)^2]d(cosx)
=∫√[1+(tant)^2][1/(cost)^2]dt
=∫[1/(cost)^3]dt
=∫[cost/(cost)^4]dt
=∫[1/(cost)^4]d(sint)
=∫{1/[1-(sint)^2]^2}d(sint)。
令sint=y,则:
∫√[1+(cosx)^2]d(cosx)
=∫[1/(1-y^2)^2]dy
=(1/4)∫[1/(1-y)+1/(1+y)]^2dy
=(1/4)∫[1/(1-y)^2]dy+(1/2)∫[1/(1-y^2)]dy+(1/4)∫[1/(1+y)^2]dy
=(1/4)/(1-y)-(1/4)/(1+y)+(1/4)∫[1/(1-y)+1/(1+y)]dy
=(1/4)[(1+y)-(1-y)]/(1-y^2)-(1/4)ln|1-y|+ln|1+y|+C
=(1/2)y/(1-y^2)+(1/4)ln|(1+y)/(1-y)|+C
=(1/2)sint/[1-(sint)^2]+(1/4)|(1+sint)/(1-sint)|+C
=(1/2)tant/cost+(1/4)ln|(1+sint)^2/(cost)^2|+C
=(1/2)u√[1+(tant)^2]+(1/2)ln[(1/cost+tant]+C
=(1/2)u√(1+u^2)+(1/2)ln{√[1+(tant)^2]+u}+C
=(1/2)cosx√[1+(cosx)^2]+(1/2)ln{√[1+(cosx)^2]+cosx}+C。
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