#int(sec(x))dx=int1/cos(x) dx#
#=intcos(x)/(cos²(x))dx#
#=intcos(x)/(1-sin²(x))dx#
Now let's substitue:
#t=sin(x)#
#dt=cos(x)dx#
#=int1/(1-t²)dt#
Now let's simplify #1/(1-t²)#
#1/(1-t²)=1/((1-t)(1+t))=A/(1-t)+B/(1+t)#
So :#(A+B)t+A-B=1#
So, by identification:
#A-B=1#(1)
#A+B=0#(2)
(1)=(1)+(2)
#2A=1#(1)
#A+B=0#(2)
So : #A=1/2, B=-1/2#
So : #1/(1-t²)=1/(2(1-t))-1/(2(1+t))#
So : #intdt/(1-t²)=intdt/(2(1-t))-intdt/(2(1+t))=-1/2(intdt/(1+t)-intdt/(1-t))#
So: #intdt/(1-t²)=-1/2ln((|t+1|)/(|t-1|))#
So finally :
#intsec(x)dx=-1/2ln((|sin(x)+1|) /(|sin(x)-1|))+C#, #C in RR#
By this similar way, #intcsc(x) dx=-1/2ln((|cos(x)+1|)/(|cos(x)-1|))+C#,#C in RR#.
\0/ here's our answer !
