Given,1-1e𝜃 + sin 𝜃 d𝜃 =1-1e𝜃 d𝜃 +1-1sin 𝜃 d𝜃 =e𝜃1-1-[cos 𝜃]1-1=e1- e-1-(cos (1)- cos (-1))=e1- e-1- cos (1)+ cos (1)=e - e-1
Problem 8 :
2eecos x -1x dx
Solution :
Given,2eecos x -1x dx=cos x -1x dx=cos x dx -1x dx=[sin x - In |x|]2ee=[(sin (2e)- In|2e|)-(sin e - In|e|]= sin (2e)- In|2e|- sin e + In|e|= sin 2e - sin e + In |e||2e|= sin 2e - sin e + In 12= sin 2e - sin e - In 2