How do you integrate int 3 * (csc(t))^2/cot(t) dt?

2 Answers
Mar 23, 2016

Use a u-substitution to get -3lnabs(cot(t))+C.

Explanation:

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.

Mar 23, 2016

3ln|csc 2t -cot 2t|+const.=3ln|tan t|+const.

Explanation:

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.