How do you find the indefinite integral of #int csc^2t/cott dt#?

Question

William Abdalla · Accepted Answer

# int csc^2t/cott \ dt = ln |Atant| #

Using the trig identity #1 + cot^2theta -= csc^2theta# we have:

# int csc^2t/cott \ dt = int (1+cot^2t)/cott \ dt# # " " = int 1/cott+cot^2t/cott \ dt# # " " = int tant+cott \ dt#

Then using the standard results:

# int tanx \ dt = ln |secx| " "(+c)# # int cosx \ dt = ln |sinx| " "(+c)#

We get the solution where (#A=#constant):

Ezekiel Alexander · Answer

# -ln|cott| + C#

Start by rewriting this integral in simpler trig functions (with respect to sine and cosine). We use the identities #csctheta = 1/sintheta# and #cottheta = costheta/sintheta# to accomplish this.

#=> int(1/sin^2t)/(cost/sint) dt#

#=> int(1/sin^2t * sint/cost) dt#

#=> int(1/(sintcost))dt#

#=> int(1/(1/2sin2t))dt#

#=> int(2csc2t)dt#

#=> 2int(csc2t)dt#

We now let #u = 2t#.Then #du = 2dt# and #dt = 1/2du#.

#=> 2int(cscu)1/2du#

#=> int(cscu)du#

This is a trick integral to do. Expand the fraction by #cscu + cotu# to get:

#=>int((csc^2u + cotucscu)/(cscu+ cotu))dt#

Now make #v = cscu + cotu#. This means that #(dv)/(du) = -(csc^2u + cotucscu)#. So:

#=>int((csc^2u + cotucscu)/(v)) * (dv)/(-(csc^2u + cotucscu))#

#=>int(-1/v)dv#

#=> -int(1/v)#

#=> -ln|v| + C#

#=> -ln|cscu + cotu| + C#

#=> -ln|csc(2t) + cot(2t)| + C#

Which can be simplified to #-ln|cott| + C#

Hopefully this helps!

Ezekiel Alexander · Answer

undefined To find the indefinite integral of $\int \csc^2(t)/\cot(t) \, dt$, we can use a substitution method. Let $u = \cot(t)$. Then, $du = -\csc^2(t) \, dt$. Rewriting the integral in terms of $u$, we get: $\int -\frac{1}{u} \, du$. Integrating $-\frac{1}{u}$ with respect to $u$ gives $-\ln|u| + C$, where $C$ is the constant of integration. Substituting back $u = \cot(t)$, we have $-\ln|\cot(t)| + C$ as the indefinite integral of $\int \csc^2(t)/\cot(t) \, dt$.