# Thread: long trigo integral:

1. ## long trigo integral:

$\int {\csc^5{x}}dx$

Solutions:
$\int {\csc^3{x}}(1+\cot^2{x})dx$

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

solving for int of $\int {\csc^3{x}}dx$
$\frac{1}{2}(-\cot{x}\csc{x} + \ln |\csc{x} - \cot{x}|)$

this is where i got stuck for solving for $\int {\cot^2{x}}{\csc^3{x}}dx$ :

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

$\int {\csc^5{x} - \csc^3{x} }dx$
which is repeating the csc^5{x}

help with hints

thanks

2. $\int \cot^{2}x \ \csc^{3}x \ dx =$ $\int (\cot^{2}x )(1+\cot^{2}x) \csc x \ dx$

So $\int \cot^{2}x \csc x \ dx + \int \cot^{4}x \ \csc x \ dx$

3. now confused with the integral of
$\int \cot^{4}x \ \csc x \ dx
$

4. Originally Posted by ^_^Engineer_Adam^_^
now confused with the integral of
$\int \cot^{4}x \ \csc x \ dx
$

I don't know about anyone else, but I typically convert everything to sine and cosine functions:
$\int cot^{4}(x)csc(x)dx = \int \frac{cos^4(x)}{sin^5(x)} dx$

Do this by parts:
$u = cos(x) \implies du = sin(x) dx$

$dv = \frac{cos^3(x)}{sin^5(x)} dx$

We need v.
$v = \int \frac{cos^3(x)}{sin^5(x)} dx$

$v = \int \frac{cos^2(x)}{sin^5(x)} cos(x) dx$

Let $z = sin(x) \implies dz = cos(x) dx$
$v = \int \frac{cos^2(x)}{sin^5(x)} cos(x) dx = \int \frac{1 - z^2}{z^5} dz = \int \frac{1}{z^5} dz - \int \frac{1}{z^3}$

$v = -\frac{1}{4z^4} - \frac{1}{2z^2} = - \left ( \frac{1}{4sin^4(x)} + \frac{1}{2sin^2(x)} \right )$

So back to the original integral:
$\int cot^{4}(x)csc(x)dx = \int \frac{cos^4(x)}{sin^5(x)} dx$

$= - cos(x) \cdot \left ( \frac{1}{4sin^4(x)} + \frac{1}{2sin^2(x)} \right ) + \int \left ( \frac{1}{4sin^4(x)} + \frac{1}{2sin^2(x)} \right ) \cdot sin(x) dx$

Can you go from here?

-Dan

5. Originally Posted by ^_^Engineer_Adam^_^
$\int {\csc^5{x}}dx$
Here is a reduction formula, using integration by parts, which can be found in a stantard table of integrals.

6. thank u!