How do I evaluate $\int \frac{{csc}^{2} x}{{cot}^{3} x} d x$ $int(csc^2x)/(cot^3x) dx$ ?

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How do I evaluate $\int \frac{{csc}^{2} x}{{cot}^{3} x} d x$ $int(csc^2x)/(cot^3x) dx$ ?

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Answer 1 · 2015-03-03T23:38:53

You can write:
$\int \frac{1}{{sin}^{2} (x)} \cdot \frac{{sin}^{3} (x)}{{cos}^{3} (x)} d x = \int \frac{sin (x)}{{cos}^{3} (x)} d x =$ $int1/sin^2(x)*sin^3(x)/cos^3(x)dx=intsin(x)/cos^3(x)dx=$
You can now use the fact that: $d [cos (x)] = - sin (x) d x$ $d[cos(x)]=-sin(x)dx$
Your integral becomes:
$\int - \frac{d [cos (x)]}{{cos}^{3} (x)} = \int - {cos}^{- 3} (x) d [cos (x)] = \frac{1}{2 \cdot {cos}^{2} (x)} + c$ $int-(d[cos(x)])/cos^3(x)=int-cos^(-3)(x)d[cos(x)]=1/(2*cos^2(x))+c$

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

Answer 2 · 2015-08-12T01:56:48

You can use the identity:

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

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

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

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

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

${csc}^{2} x = \frac{{sin}^{2} x + {cos}^{2} x}{1 - {cos}^{2} x}$ $color(green)(csc^2x) = (sin^2x + cos^2x)/(1-cos^2x)$

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

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

$= 1 + {cot}^{2} x$ $= color(green)(1 + cot^2x)$

Anyways, you get:

$\int \frac{{csc}^{2} x}{{cot}^{3} x} d x$ $int csc^2x/(cot^3x)dx$

$= \int \frac{1 + {cot}^{2} x}{{cot}^{3} x} d x$ $= int (1+cot^2x)/(cot^3x)dx$

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

$= \int (tan x) ({tan}^{2} x + 1) d x$ $= int (tanx)(tan^2x + 1) dx$

EGADS! Another identity!

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

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

$u = tan x$ $u = tanx$
$d u = {sec}^{2} x d x$ $du = sec^2xdx$

$\Rightarrow \int u d u = \frac{u^{2}}{2} + C$ $=> int udu = u^2/2 + C$

$= \frac{{tan}^{2} x}{2} + C$ $= color(blue)(tan^2x/2 + C)$

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

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

along with a domain/constraints concern.

Answer 3 · 2015-08-12T19:11:01

Here is yet a third method of evaluation.

Explanation:

$\int \frac{{csc}^{2} x}{{cot}^{3} x} d x = \int {(cot x)}^{- 3} {csc}^{2} x d x$ $int csc^2x/cot^3x dx = int(cotx)^-3 csc^2x dx$

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

$- \int u^{- 3} d u = - \frac{u^{- 2}}{- 2} + C$ $-int u^-3 du = -u^-2 /(-2) +C$

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

$= \frac{{(cot x)}^{- 2}}{2} + C$ $= (cotx)^-2/2 +C$

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

Answer 4 · 2015-08-12T19:14:50

And a fourth.

Explanation:

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

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

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

$= {(cos x)}^{- 3} sin x$ $= (cosx)^-3 sinx$

So the integral becomes:

$\int {(cos x)}^{- 3} sin x d x$ $int (cosx)^-3 sinx dx$

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

$- \frac{{(cos x)}^{- 2}}{- 2} + C$ $-(cosx)^-2/-2 +C$

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

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How do I evaluate $\int \frac{{csc}^{2} x}{{cot}^{3} x} d x$ $int(csc^2x)/(cot^3x) dx$ ?

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