Question

How do you find the integral of $\dfrac{1}{{{{\sin }^2}(x)}}$?

Answer 1

Here, we will first simplify the given integrand into an integrable form. We will divide the numerator and denominator by a suitable trigonometric ratio and simplify it. Then, we will use the substitution method and formula of integration to find the required value.

Formula used:
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$, where $n \ne - 1$

Complete step by step solution:
We have to find the value of
$\int {\dfrac{1}{{{{\sin }^2}(x)}}} dx$ ………$\left( 1 \right)$
Let us convert the integrand into an integrable form.
We do this by dividing both the numerator and denominator by the trigonometric ratio ${\cos ^2}(x)$. On doing so, we get the numerator as $\dfrac{1}{{{{\cos }^2}(x)}}$ and the denominator as $\dfrac{{{{\sin }^2}(x)}}{{{{\cos }^2}(x)}}$.
So, equation $\left( 1 \right)$ becomes
$\Rightarrow \int {\dfrac{1}{{{{\sin }^2}(x)}}} dx = \int {\dfrac{{\left( {\dfrac{1}{{{{\cos }^2}(x)}}} \right)}}{{\left( {\dfrac{{{{\sin }^2}(x)}}{{{{\cos }^2}(x)}}} \right)}}} dx$ ……..$\left( 2 \right)$
We know that the reciprocal of the ratio $\cos (x)$ is the ratio $\sec (x)$. We also know that $\tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}}$.
Using this we can write equation $\left( 2 \right)$ as:
$\Rightarrow \int {\dfrac{1}{{{{\sin }^2}(x)}}} dx = \int {\dfrac{{{{\sec }^2}(x)}}{{{{\tan }^2}(x)}}dx}$ ………$\left( 3 \right)$
We will use the substitution method in equation $\left( 3 \right)$ to simplify the integration.
For this, let us take the trigonometric ratio in the denominator of the integrand to be some variable i.e.,
$u = \tan (x)$ ………$\left( 4 \right)$
We will now differentiate the above equation on both sides. Therefore, we get
$\Rightarrow du = d(\tan (x)) = {\sec ^2}(x)dx$ ………$\left( 5 \right)$
We see that the derivative of $\tan (x)$ is the numerator of the integrand in equation $\left( 3 \right)$.
Substituting equation $\left( 4 \right)$ and $\left( 5 \right)$ in equation $\left( 3 \right)$, we have
$\int {\dfrac{1}{{{{\sin }^2}(x)}}} dx = \int {\dfrac{1}{{{u^2}}}du}$
$\Rightarrow \int {\dfrac{1}{{{{\sin }^2}(x)}}} dx = \int {{u^{ - 2}}du}$
To integrate the RHS, we will use the integration formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$, where $x = u$ and $n = - 2$. So, we get the RHS as
$\int {{u^{ - 2}}du} = \dfrac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}}$
$\Rightarrow \int {{u^{ - 2}}du} = - \dfrac{1}{u}$
From equation $\left( 4 \right)$, we have $u = \tan (x)$.Thus, the integral becomes
$\int {\dfrac{1}{{{{\sin }^2}(x)}}} dx = - \dfrac{1}{{\tan (x)}} + c$
But we know that the reciprocal of the ratio $\tan (x)$ is $\cot (x)$. Therefore, we get the value of the required integral as

$\Rightarrow \int {\dfrac{1}{{{{\sin }^2}(x)}}} dx = - \cot (x) + c$

Note:
An alternate method to find the value of the above integral would be to use the reciprocal of the ratio $\sin (x)$ and then to apply the direct integration formula. The reciprocal of $\sin (x)$ is $\csc (x)$
$\therefore \dfrac{1}{{{\sin }^{2}}(x)}={{\csc }^{2}}(x)$
Integrating both sides, we get
$\Rightarrow \int {\left( {\dfrac{1}{{{{\sin }^2}(x)}}} \right)} = \int {{{\csc }^2}(x)}$
The integration formula for ${\csc ^2}(x)$ is $\int {{{\csc }^2}(x)dx} = - \cot (x) + c$. Using this formula in above equation, we get
$\Rightarrow \int {\dfrac{1}{{{{\sin }^2}(x)}}} = - \cot (x) + c$
This value is the same value as we have obtained by the above method.