How do you find the integral of #int 1/(1 + cot(x))#?

Question

Aurora Alvarado · Accepted Answer

#1/2x-1/2ln|sinx+cosx|+C#.

Let #I=int1/(1+cotx)dx=intsinx/(sinx+cosx)dx#. Recall that, #d/dx(sinx+cosx)=cosx-sinx,# so that, #I=1/2int(2sinx)/(sinx+cosx)dx# #=1/2int{(sinx+cosx)-(cosx-sinx)}/(sinx+cosx)dx# #=1/2int(sinx+cosx)/(sinx+cosx)dx-1/2int(cosx-sinx)/(sinx+cosx)dx# #=1/2int1dx-1/2int{d/dx(sinx+cosx)}/(sinx+cosx)dx# #:. I=1/2x-1/2ln|sinx+cosx|+C#. Keep in mind that the later integral was obtained as a particular instance of #"the Result : "int(f'(x))/f(x)dx=ln|f(x)+c.# This useful Result can easily be proved by substituting #f(x)=t.# Have fun with math!

Isaiah Allen · Answer

#-1/2(ln(abs(sin(x)+cos(x)))-x)+C#

Writing this with all tangents is another way to do it:

#I=int1/(1+cot(x))dx=inttan(x)/(tan(x)(1+cot(x)))=inttan(x)/(tan(x)+1)dx#

Now, since all we have are tangents, we need a #sec^2(x)# in order to substitute. We can multiply the fraction by #sec^2(x)# in the numerator, but express it as #tan^2(x)+1# in the denominator (they're equal through the Pythagorean identity).

#I=int(tan(x)sec^2(x))/((tan(x)+1)(tan^2(x)+1))dx#

Letting #u=tan(x)# so that #du=sec^2(x)dx#, we see that:

#I=intu/((u+1)(u^2+1))du#

Currently, partial fraction decomposition needs to be done:

#u/((u+1)(u^2+1))=A/(u+1)+(Bu+C)/(u^2+1)#

Increasing by:

#u=A(u^2+1)+(Bu+C)(u+1)#

#u=Au^2+A+Bu^2+Bu+Cu+C#

Factor in three groups: those with #u^2#, those with #u#, and constants.

#u=u^2(A+B)+u(B+C)+(A+C)#

#color(purple)0u^2+color(red)1u+color(brown)0=u^2color(purple)((A+B))+ucolor(red)((B+C))+color(brown)((A+C))#

When we contrast the two, we observe that:

#{(A+B=0),(B+C=1),(A+C=0):}#

Subtracting the second equation from the third, we see that #A-B=-1#. Adding this to the first equation shows that #2A=-1# and #A=-1/2#. It then follows that:

#{(A=-1/2),(B=1/2),(C=1/2):}#

So:

#u/((u+1)(u^2+1))=1/2(1/(u+1))+1/2((u+1)/(u^2+1))#

Now let's go back to the integral:

#I=-1/2int1/(u+1)du+1/2int(u+1)/(u^2+1)du#

#I=-1/2int1/(u+1)du+1/2intu/(u^2+1)du+1/2int1/(u^2+1)du#

A small adjustment to the second integral:

#I=-1/2int1/(u+1)du+1/4int(2u)/(u^2+1)du+1/2int1/(u^2+1)du#

It is now fairly easy to integrate all three integrals:

#I=-1/2ln(abs(u+1))+1/4ln(abs(u^2+1))+1/2arctan(u)#

#I=-1/2ln(abs(tan(x)+1))+1/4ln(tan^2(x)+1)+1/2arctan(tan(x))#

#color(blue)(I=-1/2ln(abs(tan(x)+1))+1/4ln(sec^2(x))+1/2x#

After the integration constant is added, this is a good final solution, but we can play around a bit more to get some entertaining simplification.

#I=-1/2ln(abs((sin(x)+cos(x))/cos(x)))+1/2(1/2ln(sec^2(x)))+1/2x#

Rather sneakily, bring one of the #1/2#s outside the #ln(sec^2(x))# in, effectively using the #log(a^b)=blog(a)# rule in reverse. (Absolute value bars will be added since we've just taken the square root:)

#I=-1/2ln(abs((sin(x)+cos(x))/cos(x)))+1/2ln(abs(sec(x)))+1/2x#

Now we can bring a #-1# in as a #-1# power to make #sec(x)# into #cos(x)#:

#I=-1/2ln(abs((sin(x)+cos(x))/cos(x)))-1/2ln(abs(cos(x)))+1/2x#

Factor #-1/2# and use the rule that #log(A)+log(B)=log(AB)#:

#I=-1/2(ln(abs((sin(x)+cos(x))/cos(x)))+ln(abs(cos(x)))-x)#

#color(green)(I=-1/2(ln(abs(sin(x)+cos(x)))-x)+C#

Axel Alvarez · Answer

undefined To find the integral of $\int \frac{1}{1 + \cot(x)}$, you can start by using the trigonometric identity $\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}$. Then, substitute $\cot(x)$ with $\frac{\cos(x)}{\sin(x)}$. After that, use the substitution $u = \sin(x)$ and $du = \cos(x)dx$. This transforms the integral into a standard form that can be integrated easily.