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Integral of cosecant

The integral formula of csc is:

\displaystyle \int \csc u \cdot du = -\log\left( \cot u + \csc u\right)

Let’s see some examples for integrals of cosecant:

Example 1. Integral of csc x

\displaystyle \int \csc x \ dx

The cosecan integral can be deduced if you know some derivation formulas, then we will see how to get to the result of -\log \left(\cot u+\csc u \right) with the derivative formulas. Let us begin!

To start with this integral, there is a very powerful trick that allows us to solve it, we will multiply the \csc x for a fraction that contains \csc x + \cot x in the denominator and in the numerator.

\displaystyle \int \csc x \cfrac{\csc x + \cot x}{\csc x + \cot x} dx

We will do the multiplication and the obtained result will be the following:

\displaystyle \int \cfrac{\csc^{2}x + \csc x \cot x}{\csc x + \cot x} dx

Next we will name u to \csc x + \cot x and we will derive. Remember the derivation formula that say that the derivative of \csc x equals -\csc x \cot x and the derivative of \cot x equals -\csc^{2}x:

u = \csc x + \cot x

du = \left(-\csc x \cot x - \csc^{2} x\right) dx

Factoring the negative sign, it would be as follows:

du = - \left( \csc x \cot x + \csc^{2} x \right) dx

Once we have u and du, we will substitute them in the integral, as follows:

\displaystyle \int -\cfrac{1}{u} \ du

Applying properties of the integrals, we get the negative of the integral:

\displaystyle -\int\cfrac{1}{u} \ du

And now applying directly an integration formula, we will obtain the following result of our integral:

\displaystyle -\int \cfrac{1}{u} \ du = -\log (u)

Finally we substitute the u to obtain our final result:

-\log\left(\csc x + \cot x\right)

Example 2. Integral of cosecant of 2x

\displaystyle \int \csc (2x) dx

The resolution process is very similar to the previous example, let’s start.

Let’s start replacing 2x for v, then we derive and clear:

v = 2x

dv = 2 \ dx

\cfrac{dv}{2} =  dx

We substitute the integral 2x for v and dx for \frac{dv}{2}:

\displaystyle \int \csc v  \cfrac{dv}{2}

Applying properties of the integrals we will extract the denominator 2 of the integral:

\displaystyle \cfrac{1}{2} \int \csc v  \ dv

From here we can forget the \frac{1}{2} (obviously we will not forget the \frac{1}{2}) and do the integration directly with formula or apply the whole procedure already explained in the previous example.

Let’s multiply the \csc v by a fraction containing \csc v + \cot v in the numerator and in the denominator.

\displaystyle \cfrac{1}{2}\int \csc v \cfrac{\csc v + \cot v}{\csc v + \cot v} dv

We will do the multiplication and the obtained result will be the following:

\cfrac{1}{2} \int \cfrac{\csc^{2}v + \csc v \cot v}{\csc v + \cot v} dv

Next we will name u a \csc v+\cot v and we will derive. Remember the derivation formula that say that the derivative of \csc x equals -\csc x\cot x and the derivative of \cot x equals -\csc^{2}x:

u = \csc v + \cot v

du = \left(-\csc v \cot v - \csc^{2} v\right) dv

We will factor the negative:

du = - \left( \csc v \cot v + \csc^{2} v \right) dv

Once we have u and du, we will substitute them in the integral, as follows:

\displaystyle \cfrac{1}{2} \int -\cfrac{1}{u} \ du

Applying properties of integrals we will extract the negative of the integral:

\displaystyle -\cfrac{1}{2} \int \cfrac{1}{u} \ du

And now applying directly an integration formula, we will obtain the following result of our integral:

\displaystyle -\cfrac{1}{2} \int \cfrac{1}{u} \ du = -\log (u)

We substitute the u to obtain:

-\cfrac{1}{2} \log\left(\csc v + \cot v \right)

And finally we substitute v for 2x to obtain our final result:

-\cfrac{1}{2} \log \left(\csc 2x + \cot 2x\right)

Example 3. Integral of square cosecant

\displaystyle \int \csc^{2}x \ dx

Maybe some of them cause a bit of conflict the resolution of this integral, but we know very well that the integrals and the derivatives are linked by a very strong friendship bond, which means that if someone derives something and the result of that something is integrated, the result will be that something with which it started. If someone derives x, he will get 1, if he integrates 1, he will get x.

The derivative of \cot x is equal to -\csc^{2}x \ dx, that means that the integral of \csc^{2}x\ dx is -\cot x, so our final end result of \csc^{2}x \ dx will be:

-\cot x

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