7/23/2019 Integral Cscx

http://slidepdf.com/reader/full/integral-cscx 1/1

Integral of csc(x):

   1sin(x)

dx=

   sin(x)

sin2(x)dx=

   sin(x)1− cos2(x)

dx

Now use u - substitution and sub in u for cos(x). Then use partial fractions to get...

   11−u2

du=  − 12

    11−u

 +  11+u

du=−

 12(ln|u+1| − ln|u− 1|)+C  =

 12

ln

u− 1u+1

+C 

To get the final anser substitute cos(x) back in for u, and use some properties of logarithms tosimplify.

12

ln

cos(x)− 1cos(x)+1

+C  = 12

ln

cos(x)− 1cos(x)+ 1

cos(x)− 1cos(x)− 1

+C  = 12

ln

(cos(x)− 1)2

1− cos2(x)

+C 

= 12

ln

(1− cos(x))

sin(x)

2

+C  = 2 12

ln

(1− cos(x))

sin(x)

+C  = ln

1

sin(x) −

 cos(x)sin(x)

+C 

=   ln|csc(x) − cot(x)|+C 

Alternative solution (somewhat easier):

   csc(x)dx=

  csc(x)

csc(x)− cot(x)csc(x)− cot(x)

dx=

  csc2(x)− cot(x) csc(x)

csc(x)− cot(x)

dx

Use u substtition with u = csc(x) - cot(x) to get...

   du

u  = ln|u|= ln|csc(x)− cot(x)|+C 

Notice this is the same equation obtained before.

1