求如图不定积分 20
2个回答
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命令x＝at进行换元
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∫dx/(x+√a^2-x^2)
=∫d(x/a)/(x/祥世a+√(1-(x/a)^2) )
令x/a=sinu
=∫dsinu/(cosu+sinu)
=(1/滑洞√2)∫cosudu/sin(u+π/4)
=(1/√2)∫[cos(u+π/4)cos(π/信宴枯4)+sin(u+π/4)sin(π/4)]du/sin(u+π/4)
=(1/2)∫cos(u+π/4)du/sin(u+π/4)+ (1/2)u
=(1/2)ln|sin(u+π/4)|+(1/2)u +C
反带回x
=(1/2)ln|√2[x/a+√(1-x^2/a^2)]|+(1/2)u+C0
=(1/2)ln|x+√(a^2-x^2)|+(1/2)arcsin(x/a)+C
=∫d(x/a)/(x/祥世a+√(1-(x/a)^2) )
令x/a=sinu
=∫dsinu/(cosu+sinu)
=(1/滑洞√2)∫cosudu/sin(u+π/4)
=(1/√2)∫[cos(u+π/4)cos(π/信宴枯4)+sin(u+π/4)sin(π/4)]du/sin(u+π/4)
=(1/2)∫cos(u+π/4)du/sin(u+π/4)+ (1/2)u
=(1/2)ln|sin(u+π/4)|+(1/2)u +C
反带回x
=(1/2)ln|√2[x/a+√(1-x^2/a^2)]|+(1/2)u+C0
=(1/2)ln|x+√(a^2-x^2)|+(1/2)arcsin(x/a)+C
追答
满意请采纳
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