Question

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

Answer 1

Hint: 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.