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

Mar 23, 2016

Use a $u$-substitution to get $- 3 \ln \left\mid \cot \left(t\right) \right\mid + C$.

#### Explanation:

First, note that because $3$ is a constant, we can pull it out of the integral to simplify:
$3 \int \frac{{\csc}^{2} \left(t\right)}{\cot} \left(t\right) \mathrm{dt}$

Now - and this is the most important part - notice that the derivative of $\cot \left(t\right)$ is $- {\csc}^{2} \left(t\right)$. Because we have a function and its derivative present in the same integral, we can apply a $u$ substitution like this:
$u = \cot \left(t\right)$
$\frac{\mathrm{du}}{\mathrm{dt}} = - {\csc}^{2} \left(t\right)$
$\mathrm{du} = - {\csc}^{2} \left(t\right) \mathrm{dt}$

We can convert the positive ${\csc}^{2} \left(t\right)$ to a negative like this:
$- 3 \int \frac{- {\csc}^{2} \left(t\right)}{\cot} \left(t\right) \mathrm{dt}$

And apply the substitution:
$- 3 \int \frac{\mathrm{du}}{u}$

We know that $\int \frac{\mathrm{du}}{u} = \ln \left\mid u \right\mid + C$, 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 $u = \cot \left(t\right)$, we can say:
$- 3 \left(\ln \left\mid u \right\mid + C\right) = - 3 \ln \left\mid \cot \left(t\right) \right\mid + C$

And that's all.

Mar 23, 2016

$3 \ln | \csc 2 t - \cot 2 t | + c o n s t . = 3 \ln | \tan t | + c o n s t .$

#### Explanation:

$3 \int {\csc}^{2} \frac{t}{\cot} t \mathrm{dt} =$
$= 3 \int \left(\frac{1}{\sin} ^ 2 t\right) \cdot \left(\frac{1}{\cos \frac{t}{\sin} t}\right) \mathrm{dt}$
$= 3 \int \frac{\mathrm{dt}}{\sin t \cdot \cos t}$

Remember that
$\sin 2 t = 2 \sin t \cdot \cos t$

So
$= 3 \int \frac{\mathrm{dt}}{\left(\frac{1}{2}\right) \sin 2 t}$
$= 6 \int \csc 2 t \cdot \mathrm{dt}$

As we can find in a table of integrals
(for instance Table of integrals containing Csc (ax) in SOS Math ):

$\int \csc a x \cdot \mathrm{dx} = \frac{1}{a} \ln | \csc a x - \cot a x | = \ln | \tan \left(\frac{a x}{2}\right) |$

we get this result
$= 3 \ln | \csc 2 t - \cot 2 t | + c o n s t = 3 \ln | \tan t | + c o n s t .$