Question

How do you integrate ${\left( {\csc 2x - \cot 2x} \right)^2}dx$?

Answer 1

Hint : Here in this question, we have to find the integrated value of a given trigonometric function. Firstly, we have to rewrite a given equation using a definition of trigonometric ratios i.e., $\csc \theta = \dfrac{1}{{\sin \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ then later by using partial fractions method. I would try Integration by Partial Fractions when an integrand is a rational function with its denominator can be factored out into smaller factors and by further simplifying using the standard integration formula. To get the required solution.

Complete step by step solution:
Consider
$\Rightarrow \,\,\int {{{\left( {\csc 2x - \cot 2x} \right)}^2}dx}$---------(1)
As we know the definition of trigonometric ratios:
Cosecant is a reciprocal of sine i.e., $\csc \theta = \dfrac{1}{{\sin \theta }}$ and
Cotangent is a ratio between cosine and sine i.e., $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
Then, equation (1) can be rewritten as
$\Rightarrow \,\,\int {{{\left( {\dfrac{1}{{\sin 2x}} - \dfrac{{\cos 2x}}{{\sin 2x}}} \right)}^2}dx}$
On simplification, we have
$\Rightarrow \,\,\int {{{\left( {\dfrac{{1 - \cos 2x}}{{\sin 2x}}} \right)}^2}dx}$
Or
$\Rightarrow \,\,\int {\dfrac{{{{\left( {1 - \cos 2x} \right)}^2}}}{{{{\sin }^2}2x}}dx}$
By the trigonometric ratio ${\sin ^2}\theta + {\cos ^2}\theta = 1 \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta$, then
$\Rightarrow \,\,\int {\dfrac{{{{\left( {1 - \cos 2x} \right)}^2}}}{{1 - {{\cos }^2}2x}}dx}$
In denominator we observed a identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
Here, $a = 1$ and $b = \cos 2x$, then
$\Rightarrow \,\,\int {\dfrac{{{{\left( {1 - \cos 2x} \right)}^2}}}{{\left( {1 - \cos 2x} \right)\left( {1 + \cos 2x} \right)}}dx}$
On simplification, we get
$\Rightarrow \,\,\int {\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}dx}$--------(2)
Now, the integrand (the expression after the integral sign) is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be expressed in partial fractions before integration takes place.
The steps needed to decompose an algebraic fraction into its partial fractions results from a consideration of the reverse process − addition (or subtraction).
Separate the fraction that we wish to decompose in to multiple fractions. The factor of $x$ in the denominator has a power higher than 1, then the coefficients in the numerator should reflect this higher power.
To use the Method of Partial Fractions, we let, for, $\;A,B \in R,:$
The function of equation (2) can be written as
$\Rightarrow \,\,\,\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}} = \dfrac{A}{{1 + \cos 2x}} + B$--------(3)
Take $\left( {1 + \cos 2x} \right)$ as LCM in RHS, then
$\Rightarrow \,\,\,1 - \cos 2x = A + B\left( {1 + \cos 2x} \right)$
$\Rightarrow \,\,\,1 - \cos 2x = A + B + B\cos 2x$----------(4)
By comparing the constant coefficient on both side, we have
$\Rightarrow \,\,\,A + B = 1$
By comparing the $\cos 2x$ co-efficient on both side, we have
$\Rightarrow \,\,\,B = - 1$
$\therefore \,\,\,B = - 1$
Then $A$ value is:
$\Rightarrow \,\,\,A + - 1 = 1$
$\Rightarrow \,\,\,A = 1 + 1$
$\therefore \,\,\,A = 2$
Substitute the value of $A$ and $B$in equation (3):
$\Rightarrow \,\,\,\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}} = \dfrac{2}{{1 + \cos 2x}} + \left( { - 1} \right)$
$\Rightarrow \,\,\,\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}} = \dfrac{2}{{1 + \cos 2x}} - 1$
Integrate each fraction with respect to $x$
$\Rightarrow \,\,\,\int {\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}\,dx} = \int {\dfrac{2}{{1 + \cos 2x}}\,dx} - \int {1\,dx}$----(5)
By integrating $\Rightarrow \,\,\,\int {\dfrac{2}{{1 + \cos 2x}}\,dx}$ we get
By using the trigonometric ratio ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $\,\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$, then
$\Rightarrow \,\,\,\int {\dfrac{2}{{{{\sin }^2}x + {{\cos }^2}x + \,{{\cos }^2}x - {{\sin }^2}x}}\,dx}$
On simplification, we have
$\Rightarrow \,\,\,\int {\dfrac{2}{{2{{\cos }^2}x}}\,dx}$
$\Rightarrow \,\,\,\int {\dfrac{1}{{{{\cos }^2}x}}\,dx}$
By using the definition of trigonometric ratio $\sec \theta = \dfrac{1}{{\cos \theta }}$, then
$\Rightarrow \,\,\,\int {{{\sec }^2}x\,dx}$
On integrating, we get
$\Rightarrow \,\,\,\tan x + {c_1}$-------(a)
Next, integrate
$\Rightarrow \,\,\,\int {1\,dx}$
On integrating, we get
$\Rightarrow \,\,\,x + {c_2}$-------(b)
Hence, substitute the (a) and (b) in equation (5) to get the required solution
$\Rightarrow \,\,\,\int {\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}\,dx} = \tan x + {c_1} - x - {c_2}$
Where, ${c_1}$ and ${c_2}$ are integrating constant which is equal to $C$
$\Rightarrow \,\,\int {{{\left( {\csc 2x - \cot 2x} \right)}^2}dx} = \,\int {\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}\,dx} = \tan {x} - x + C$
Where $C$ is an integrating constant.
So, the correct answer is “$\int {\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}\,dx} = \tan {x} - x + C$”.

Note : When the function is in the form of a fraction where the denominator of the function is in the form of a polynomial, then we use the partial fractions and then we integrate the function. The integration is a reciprocal of the differentiation and by using the standard formulas of integration we determine the values.