CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If  f(x)=ex(21tanx+tan2(x+π4)) dx, where f(3π4)=0. Then the value of  ln(f(π)) is 


Solution

f(x)=ex(1+tan x1tanx+sec2(x+π4)) dx
=ex(tan(π4+x)+sec2(x+π4)) dx
f(x)=ex tan(π4+x)+C
f(3π4)=0C=0
f(π)=eπln(f(x))=π

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More...



footer-image