# How do I evaluate int(csc^2x)/(cot^3x) dx?

Mar 3, 2015

You can write:
$\int \frac{1}{\sin} ^ 2 \left(x\right) \cdot {\sin}^{3} \frac{x}{\cos} ^ 3 \left(x\right) \mathrm{dx} = \int \sin \frac{x}{\cos} ^ 3 \left(x\right) \mathrm{dx} =$
You can now use the fact that: $d \left[\cos \left(x\right)\right] = - \sin \left(x\right) \mathrm{dx}$
$\int - \frac{d \left[\cos \left(x\right)\right]}{\cos} ^ 3 \left(x\right) = \int - {\cos}^{- 3} \left(x\right) d \left[\cos \left(x\right)\right] = \frac{1}{2 \cdot {\cos}^{2} \left(x\right)} + c$

Where you used $\cos \left(x\right)$ as if it was $x$ in a normal integral integrating ${\cos}^{- 3} \left(x\right)$ as if it was ${x}^{- 3}$

Aug 12, 2015

You can use the identity:

${\csc}^{2} x = 1 + {\cot}^{2} x$

If you don't remember that, you can derive it like so:

${\sin}^{2} x + {\cos}^{2} x = 1$

${\sin}^{2} x = 1 - {\cos}^{2} x$

$\frac{1}{\sin} ^ 2 x = \frac{1}{1 - {\cos}^{2} x}$

$\textcolor{g r e e n}{{\csc}^{2} x} = \frac{{\sin}^{2} x + {\cos}^{2} x}{1 - {\cos}^{2} x}$

$= \frac{{\sin}^{2} x + {\cos}^{2} x}{{\sin}^{2} x}$

$= 1 + \frac{{\cos}^{2} x}{{\sin}^{2} x}$

$= \textcolor{g r e e n}{1 + {\cot}^{2} x}$

Anyways, you get:

$\int {\csc}^{2} \frac{x}{{\cot}^{3} x} \mathrm{dx}$

$= \int \frac{1 + {\cot}^{2} x}{{\cot}^{3} x} \mathrm{dx}$

$= \int {\tan}^{3} x + \tan x \mathrm{dx}$

$= \int \left(\tan x\right) \left({\tan}^{2} x + 1\right) \mathrm{dx}$

$= \int \tan x {\sec}^{2} x \mathrm{dx}$

Now we can just do a bit of quick u-substitution. Let:

$u = \tan x$
$\mathrm{du} = {\sec}^{2} x \mathrm{dx}$

$\implies \int u \mathrm{du} = {u}^{2} / 2 + C$

$= \textcolor{b l u e}{{\tan}^{2} \frac{x}{2} + C}$

Now the weird part is, up until and including the ${u}^{2} / 2$, it seems to be correct. But Wolfram Alpha says it is ${\sec}^{2} \frac{x}{2} + C$. If I had to guess, I would say it was because:

${\tan}^{2} \frac{x}{2} + C \propto {\tan}^{2} \frac{x}{2} + \frac{1}{2} + C \propto {\sec}^{2} \frac{x}{2} + C$

along with a domain/constraints concern.

Aug 12, 2015

Here is yet a third method of evaluation.

#### Explanation:

$\int {\csc}^{2} \frac{x}{\cot} ^ 3 x \mathrm{dx} = \int {\left(\cot x\right)}^{-} 3 {\csc}^{2} x \mathrm{dx}$

Let $u = \cot x$ so that $\mathrm{du} = - {\csc}^{2} x \mathrm{dx}$ and the integral becomes:

$- \int {u}^{-} 3 \mathrm{du} = - {u}^{-} \frac{2}{- 2} + C$

$= {u}^{-} \frac{2}{2} + C$

$= {\left(\cot x\right)}^{-} \frac{2}{2} + C$

$= \frac{1}{2 {\cot}^{2} x} + C = {\tan}^{2} \frac{x}{2} + C$

Aug 12, 2015

And a fourth.

#### Explanation:

${\csc}^{2} \frac{x}{\cot} ^ 3 x = {\tan}^{3} x {\csc}^{2} x$

$= {\sin}^{3} \frac{x}{\cos} ^ 3 x \frac{1}{\sin} ^ 2 x$

$= \sin \frac{x}{\cos} ^ 3 x$

$= {\left(\cos x\right)}^{-} 3 \sin x$

So the integral becomes:

$\int {\left(\cos x\right)}^{-} 3 \sin x \mathrm{dx}$

which can be evaluated using $u = \cos x$ to give:

$- {\left(\cos x\right)}^{-} \frac{2}{-} 2 + C$

$= {\sec}^{2} \frac{x}{2} + C$