Question
Thu January 24, 2013 By: Prashaant Malaviya
 

integral of 1+sinx/1-sinx?

Expert Reply
Thu January 24, 2013
?(1+sinx/1-sinx)dx
using
1+sinx=sin^2(x/2)+cos^2(x/2)+2sin(x/2)*cos(x/2)
         =[sin(x/2)+cos(x/2)]^2
1-sinx=[sin(x/2)-cos(x/2)]^2
as in the question it is square so will not affect
sin(X/2)-COS(X/2)
OR
COS(X/2)-SIN(X/2)
?[{sin(x/2)+cos(x/2)}/{cos(x/2)-sin(x/2)}]^2 dx
?{1+tan(x/2)}/{1-tan(x/2)}^2 dx
?tan^2(pi/4-x/2)dx
?{(sec^2(pi/4-x/2)-1}dx
-1/2*tan(pi/4-x/2)-x+constant
 
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