Question

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

Answer 1

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.

Answer 2

`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.`