# Thread: Integral of 1/cot(x) + 2

1. ## Integral of 1/cot(x) + 2

pls what is the integral of 1/cotx+2

thanks.

let me make it clearer guys;

the equation is dy/dx = cot(y+x)-1

solution is Cos(y+x)=ce-x

my problem is the process leading to the solution and was solved using variable saperable differencial equation

thanks for ur assistance.

2. ## Re: integration equation! pls help

You have to be more clear, do you have to calculate the integral:
$\int \frac{1}{\cot(x)+2}dx$
or
$\int \left(\frac{1}{\cot(x)}+2\right)dx$
?
To start, write $\cot(x)=\frac{\cos(x)}{\sin(x)}$

3. ## Re: integration equation! pls help

it is probably the first one because the second one is the same as inegral of tan(x)+2.

4. ## Re: integration equation! pls help

To solve the integral, I think it's better to wright: $\cot(x)=\frac{1}{\tan(x)}$.
$\int \frac{dx}{\cot(x)+2} = \int \frac{dx}{\frac{1}{\tan(x)}+2}$

Let $\tan(x)=t$ then $\frac{dx}{\cos^2(x)}=dt \Leftrightarrow dx=dt\cdot \cos^2(x)$

We want now $\cos^2(x)$ in function of $t$ so, we use the fact that: $\tan(x)=t \Leftrightarrow x=\arctan(t)$.

And $\cos^2[\arctan(t)]=\left(\frac{1}{\sqrt{1+t^2}}\right)^2=\frac{1}{1+ t^2}$

Now the integral becomes:
$\int \frac{dt}{\left(\frac{1}{t}+2\right)\cdot \left(1+t^2)}}$

Now you can try to split up in partial fractions.

5. ## Re: integration equation! pls help

Your ODE is

$\dfrac{dy}{dx} = \cot (x + y ) - 1$

If you let $u = x + y$ then

$\dfrac{du}{dx} = \cot u$

which leads to the separation

$\dfrac{du}{\cot u} = dx$ or $\tan u du = dx$.

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# integration of 1/ cot y

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