How to integrate int (secx)dx and int (cscx)dx ?

1 Answer
Apr 11, 2018

intsec(x)dx=-1/2ln((|sin(x)+1|) /(|sin(x)-1|))+C,intcsc(x) dx=-1/2ln((|cos(x)+1|)/(|cos(x)-1|))+C,C in RR

Explanation:

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 !