求积分,第二题和第三题,学到分部积分时老师留的作业,
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求积分,第二题和第三题,学到分部积分时老师留的作业,
![](http://img.wesiedu.com/upload/0/a2/0a213f2aded15e7fb7d1370348b59e93.jpg)
![](http://img.wesiedu.com/upload/0/a2/0a213f2aded15e7fb7d1370348b59e93.jpg)
![求积分,第二题和第三题,学到分部积分时老师留的作业,](/uploads/image/z/7922065-49-5.jpg?t=%E6%B1%82%E7%A7%AF%E5%88%86%2C%E7%AC%AC%E4%BA%8C%E9%A2%98%E5%92%8C%E7%AC%AC%E4%B8%89%E9%A2%98%2C%E5%AD%A6%E5%88%B0%E5%88%86%E9%83%A8%E7%A7%AF%E5%88%86%E6%97%B6%E8%80%81%E5%B8%88%E7%95%99%E7%9A%84%E4%BD%9C%E4%B8%9A%2C)
原式=-∫lnsinxdcotx=-(lnsinx cotx-∫cot^2 xdx)=-lnsinx cotx+∫1-sin^2 x/sin^2 x dx=-lnsinx cot x -cotx -x +c
令x=sinu,则:u=arcsinx,dx=cosudu. ∴∫{1/[2x+√(1-x^2)]}dx=∫[1/(2sinu+cosu)]cosudu. 引入辅助角t,使cost=2/√5、sint=1/√5.则:t=arcsin(1/√5). ∴∫{1/[2x+√(1-x^2)]}dx =∫[1/(2sinu+cosu)]cosudu =(1/√5)∫[1/(sinucost+cosusint)]cos(u+t-t)d(u+t) =(1/√5)∫{[cos(u+t)cost+sin(u+t)sint]/sin(u+t)}d(u+t) =(1/√5)cost∫[cos(u+t)/sin(u+t)]d(u+t)+(1/√5)sint∫d(u+t) =(1/√5)×(2/√5)∫[1/sin(u+t)]d[sin(u+t)]+(1/√5)×(1/√5)(u+t) =(2/5)ln|sin(u+t)|+(1/5)[arcsinx+arcsin(1/√5)]+C =(2/5)ln|sinucost+cosusint|+(1/5)arcsinx+C =(2/5)ln|(2/√5)x+(1/√5)√(1-x^2)|+(1/5)arcsinx+C =(2/5)ln|2x+√(1-x^2)|-ln√5+(1/5)arcsinx+C =(2/5)ln|2x+√(1-x^2)|+(1/5)arcsinx+C
令x=sinu,则:u=arcsinx,dx=cosudu. ∴∫{1/[2x+√(1-x^2)]}dx=∫[1/(2sinu+cosu)]cosudu. 引入辅助角t,使cost=2/√5、sint=1/√5.则:t=arcsin(1/√5). ∴∫{1/[2x+√(1-x^2)]}dx =∫[1/(2sinu+cosu)]cosudu =(1/√5)∫[1/(sinucost+cosusint)]cos(u+t-t)d(u+t) =(1/√5)∫{[cos(u+t)cost+sin(u+t)sint]/sin(u+t)}d(u+t) =(1/√5)cost∫[cos(u+t)/sin(u+t)]d(u+t)+(1/√5)sint∫d(u+t) =(1/√5)×(2/√5)∫[1/sin(u+t)]d[sin(u+t)]+(1/√5)×(1/√5)(u+t) =(2/5)ln|sin(u+t)|+(1/5)[arcsinx+arcsin(1/√5)]+C =(2/5)ln|sinucost+cosusint|+(1/5)arcsinx+C =(2/5)ln|(2/√5)x+(1/√5)√(1-x^2)|+(1/5)arcsinx+C =(2/5)ln|2x+√(1-x^2)|-ln√5+(1/5)arcsinx+C =(2/5)ln|2x+√(1-x^2)|+(1/5)arcsinx+C