Question

How do you find the integral of $\int {{{\csc }^n}\left( x \right)} $ if $m$ or $n$ is an integer?

Answer 1

Hint:As we are given $\int {{{\csc }^n}\left( x \right)} $ to solve, we will first distribute its power in such a form that we get the expression as a product of two functions. After getting the expression as a product of two numbers, we will apply the method of integration by parts i.e., $\int {udv} = uv - \int {vdu} $. We will assume one function as $u$ and the other as $dv$ and will calculate the missing terms that are, $du$ and $v$, and after obtaining all the values we will put them in the formula and simplify it.

Complete step by step solution:
(i)
Let us assume the given integrand as $I$. Therefore,
$I = \int {{{\csc }^n}\left( x \right)dx} $

As we know that we can split $n$ as a sum of $2$ and $\left( {n - 2} \right)$, therefore, we can write $I$ as:

$I = \int {{{\cos }^{\left[ {\left( {n - 2} \right) + 2} \right]}}\left( x \right)dx} $

As we know that ${a^{\left( {m + n} \right)}} = {a^m} \times {a^n}$, we can write $I$ as:
$I = \int {{{\csc }^{n - 2}}\left( x \right){{\csc }^2}\left( x \right)dx} $

(ii)
As we have got our integrand as a product of two functions, we can now perform integration by parts,

taking the form:
$\int {udv} = uv - \int {vdu} $

From our integrand, we let $u$ be ${\csc ^{n - 2}}\left( x \right)$, i.e.,
$u = {\csc ^{n - 2}}\left( x \right)$

Differentiating both sides by applying chain rule, will give us:
$du = \left( {n - 2} \right){\csc ^{n - 3}}\left( x \right) \times \left( { - \csc \left( x \right)\cot \left( x \right)} \right)dx$

On simplifying, we will get:

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

(iii)
Now, as we can see another term in our integrand $I$ is ${\csc ^2}\left( x \right)$. So, we will let $dv$

be ${\csc ^2}\left( x \right)$. Therefore,
$dv = {\csc ^2}\left( x \right)$

Integrating both the sides, we will get:
$\int {dv} = \int {{{\csc }^2}\left( x \right)dx} $

As we know that $\dfrac{d}{{dx}}\cot x = - {\csc ^2}x$ and the vice versa, $\int {{{\csc }^2}xdx} = - \cot x$ is also true

Therefore, on solving, we will get:
$v = - \cot \left( x \right)$

(iv)
Now since, we have the values of $u$, $v$, $du$ and $dv$, we will apply the formula $\int {udv} = uv - \int {vdu} $ on $I$ and it will become:
$I = - \cot \left( x \right){\csc ^{n - 2}}\left( x \right) - \left( {n - 2} \right)\int {{{\csc }^{n - 2}}\left( x
\right){{\cot }^2}\left( x \right)dx} $

Now as we know that,
$1 + {\cot ^2}x = {\csc ^2}x$

We can also write that,
${\cot ^2}x = {\csc ^2}x - 1$

So, writing ${\cot ^2}x$ as ${\csc ^2}x - 1$ in our integrand $I$, we will get:

$I = - \cot \left( x \right){\csc ^{n - 2}}\left( x \right) - \left( {n - 2} \right)\int {{{\csc }^{n - 2}}\left( x \right)\left( {{{\csc }^2}\left( x \right) - 1} \right)dx} $

On simplifying by opening the parentheses we will get:

\[I = - \cot \left( x \right){\csc ^{n - 2}}\left( x \right) - \left( {n - 2} \right)\int {\left( {{{\csc }^n}\left( x \right) - {{\csc }^{n - 2}}\left( x \right)} \right)dx} \]
(v)

As we know that $\int {(A - B)} = \int A - \int B $.

Therefore, we can write $I$ as:

\[I = - \cot \left( x \right){\csc ^{n - 2}}\left( x \right) - \left( {n - 2} \right)\int {{{\csc }^n}\left( x
\right)dx} + \left( {n - 2} \right)\int {{{\csc }^{n - 2}}\left( x \right)dx} \]

As we know that $I = \int {{{\csc }^n}\left( x \right)dx} $ as assumed in our first step, we will substitute

$\int {{{\csc }^n}\left( x \right)dx} $ as $I$:
\[I = - \cot \left( x \right){\csc ^{n - 2}}\left( x \right) - \left( {n - 2} \right)I + \left( {n - 2} \right)\int
{{{\csc }^{n - 2}}\left( x \right)dx} \]

Now, in order to shift $I$ to LHS, we will add $\left( {n - 2} \right)I$ on both the sides:
\[I + \left( {n - 2} \right)I = - \cot \left( x \right){\csc ^{n - 2}}\left( x \right) - \left( {n - 2} \right)I + \left( {n - 2} \right)\int {{{\csc }^{n - 2}}\left( x \right)dx} + \left( {n - 2} \right)I\]

On simplifying, we get:
\[\left( {n - 1} \right)I = - \cot \left( x \right){\csc ^{n - 2}}\left( x \right) + \left( {n - 2} \right)\int
{{{\csc }^{n - 2}}\left( x \right)dx} \]

Now, dividing both the sides by $\left( {n - 1} \right)$ and writing $I$ as $\int {{{\csc }^n}\left( x \right)dx} $
we will get:

$\int {{{\csc }^n}\left( x \right)dx} = \dfrac{{ - \cot \left( x \right){{\csc }^{n - 2}}\left( x \right)}}{{n - 1}} + \dfrac{{n - 2}}{{n - 1}}\int {{{\csc }^{n - 2}}\left( x \right)dx} $

Since, we have got the value of $\int {{{\csc }^n}\left( x \right)dx} $ in terms of $\int {{{\csc }^{n - 2}}\left( x \right)dx} $,

 we will repeat the process with $\int {{{\csc }^{n - 2}}\left( x \right)dx} $ until we get $\int
{{{\csc }^2}\left( x \right)dx} $ which is $ - \cot \left( x \right)$ or we get $\int {\csc \left( x
\right)dx} $ which is $ - \ln \left| {\cot \left( x \right) + \csc \left( x \right)} \right|$.

Note: Since, we are still getting $n$ in our final solution, if we are given the value of $n$ in the question and it is an even number, we will repeat the process until we get $\int {{{\csc }^2}\left( x \right)dx} $ and will write it as $ - \cot \left( x \right)$.

 Whereas if the given value of $n$ is an odd number, repeating the process will give us $\int {\csc \left( x \right)dx} $ and we will write it as $ - \ln \left| {\cot \left( x \right) + \csc \left( x \right)} \right|$.

If no value of $n$ is given in the question, we will simply write $\int {{{\csc }^n}\left( x \right)dx} = \dfrac{{ - \cot \left( x \right){{\csc }^{n - 2}}\left( x \right)}}{{n - 1}} + \dfrac{{n - 2}}{{n - 1}}\int {{{\csc }^{n - 2}}\left( x \right)dx} $.