How do you integrate #int 3 * (csc(t))^2/cot(t) dt#?
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Use a #u#-substitution to get #-3lnabs(cot(t))+C#.
First, note that because #3# is a constant, we can pull it out of the integral to simplify:
#3int(csc^2(t))/cot(t)dt#
Now - and this is the most important part - notice that the derivative of #cot(t)# is #-csc^2(t)#. Because we have a function and its derivative present in the same integral, we can apply a #u# substitution like this:
#u=cot(t)#
#(du)/dt=-csc^2(t)#
#du=-csc^2(t)dt#
We can convert the positive #csc^2(t)# to a negative like this:
#-3int(-csc^2(t))/cot(t)dt#
And apply the substitution:
#-3int(du)/u#
We know that #int(du)/u=lnabs(u)+C#, so evaluating the integral is done. We just need to reverse substitute (put the answer back in terms of #t#) and attach that #-3# to the result. Since #u=cot(t)#, we can say:
#-3(lnabs(u)+C)=-3lnabs(cot(t))+C#
And that's all.
#3ln|csc 2t -cot 2t|+const.=3ln|tan t|+const.#
#3 int csc^2 t/cot t dt=#
#=3 int (1/sin^2 t)*(1/(cos t/sin t))dt#
#=3 int dt/(sin t *cos t)#
Remember that
#sin 2t =2sint*cost#
So
#=3int dt/((1/2)sin 2t)#
#=6int csc 2t*dt#
As we can find in a table of integrals
(for instance Table of integrals containing Csc (ax) in SOS Math ):
#int csc ax*dx=1/aln|cscax-cotax|=ln|tan ((ax)/2)|#
we get this result
#=3ln|csc2t-cot2t|+const=3ln|tan t|+const.#
