Question

The value of $\int {\dfrac{{{{\cot }^2}x}}{{\left( {{{\csc }^2}x + \csc x} \right)}}dx = }$
A. $x - \sin x + c$
B. $x + \cos x + c$
C. $\sin x - x + c$
D. $x - \cos x + c$

Answer 1

We need to find the value of integral $\int {\dfrac{{{{\cot }^2}x}}{{\left( {{{\csc }^2}x + \csc x} \right)}}dx}$. For that, we will first write $\cot x$ and $\csc x$, in terms of $\sin x$ and $\cos x$. After that, we will cancel out some terms from the numerator and denominator after taking the LCM from the denominator. After all this we will be left with $\int {\dfrac{{{{\cos }^2}x}}{{1 + \sin x}}} dx$. We will then use the property ${\sin ^2}x + {\cos ^2}x = 1$ and write ${\cos ^2}x = 1 - {\sin ^2}x$which is of the form ${a^2} - {b^2}$. We know, ${a^2} - {b^2} = (a + b)(a - b)$. So, we will use this property and then our integral becomes $\int {\dfrac{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}{{\left( {1 + \sin x} \right)}}} dx$. Now, after cancelling the terms from numerator and denominator, we will solve the remaining integral to obtain the required answer.

Complete step by step answer:
We need to find $\int {\dfrac{{{{\cot }^2}x}}{{\left( {{{\csc }^2}x + \csc x} \right)}}dx}$.
Writing $\cot x$ as $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\csc x$ as $\csc x = \dfrac{1}{{\sin x}}$ in the integral, we get
$\Rightarrow \int {\dfrac{{{{\cot }^2}x}}{{\left( {{{\csc }^2}x + \csc x} \right)}}dx} = \int {\dfrac{{{{\left( {\dfrac{{\cos x}}{{\sin x}}} \right)}^2}}}{{\left( {{{\left( {\dfrac{1}{{\sin x}}} \right)}^2} + \dfrac{1}{{\sin x}}} \right)}}dx} = \int {\dfrac{{\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}}{{\left( {\dfrac{1}{{{{\sin }^2}x}} + \dfrac{1}{{\sin x}}} \right)}}dx}$
Taking LCM in the denominator, we have
$\Rightarrow \int {\dfrac{{{{\cot }^2}x}}{{\left( {{{\csc }^2}x + \csc x} \right)}}dx} = \int {\dfrac{{\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}}{{\left( {\dfrac{1}{{{{\sin }^2}x}} + \dfrac{1}{{\sin x}}} \right)}}dx} = \int {\dfrac{{\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}}{{\left( {\dfrac{{1 + \sin x}}{{{{\sin }^2}x}}} \right)}}dx}$

In simple way, we can write the integral as
$\int {\left( {\left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right) \div \left( {\dfrac{{1 + \sin x}}{{{{\sin }^2}x}}} \right)} \right)dx}$
We know, $a \div b = a \times \dfrac{1}{b}$. Using this, we get
$\Rightarrow \int {\left( {\left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right) \times \left( {\dfrac{1}{{\dfrac{{1 + \sin x}}{{{{\sin }^2}x}}}}} \right)} \right)dx}$
$\Rightarrow \int {\left( {\left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right) \times \left( {\dfrac{{{{\sin }^2}x}}{{1 + \sin x}}} \right)} \right)dx}$
Cancelling out ${\sin ^2}x$ from numerator and denominator, we get
$\Rightarrow \int {\left( {\dfrac{{{{\cos }^2}x}}{{1 + \sin x}}} \right)dx}$

Now using ${\sin ^2}x + {\cos ^2}x = 1 \Leftrightarrow {\cos ^2}x = 1 - {\sin ^2}x$ in numerator, we get
$\Rightarrow \int {\left( {\dfrac{{1 - {{\sin }^2}x}}{{1 + \sin x}}} \right)dx}$
As we know, ${(1)^2} = 1$, we can write the above equation as
$\Rightarrow \int {\dfrac{{{{\left( 1 \right)}^2} - {{\left( {\sin x} \right)}^2}}}{{1 + \sin x}}} dx$
Now using the property ${a^2} - {b^2} = (a + b)(a - b)$ in numerator, we get
$\Rightarrow \int {\dfrac{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}{{1 + \sin x}}dx}$
Cancelling out $1 + \sin x$ from numerator and denominator, we get
$\Rightarrow \int {\left( {1 - \sin x} \right)dx}$
We know, $\Rightarrow \int {\left( {f(x) - g(x)} \right)dx} = \int {f(x)} dx - \int {g(x)} dx$. So,
$\Rightarrow \int {\left( {1 - \sin x} \right)dx = \int 1 dx - \int {\sin x} dx}$

Now, using $\int 1 dx = x + c$and $\int {\sin x} dx = - \cos x + c$ in the above equation, we get
$\int {\left( {1 - \sin x} \right)dx = \int 1 dx - \int {\sin x} dx}$
$\Rightarrow \int {\left( {1 - \sin x} \right)dx} = x + {c_1} - ( - \cos x + {c_2})$, where ${c_1},{c_2}$ are constant terms
$\Rightarrow \int {\left( {1 - \sin x} \right)dx} = x + {c_1} + \cos x - {c_2}$
Clubbing the constant terms together, we get
$\int {\dfrac{{{{\cot }^2}x}}{{\left( {{{\csc }^2}x + \csc x} \right)}}dx} = x + \cos x + \left( {{c_1} - {c_2}} \right)$
$\Rightarrow \int {\dfrac{{{{\cot }^2}x}}{{\left( {{{\csc }^2}x + \csc x} \right)}}dx} = x + \cos x + c,$ where $c = {c_1} - {c_2}$
Hence, we got $\int {\dfrac{{{{\cot }^2}x}}{{\left( {{{\csc }^2}x + \csc x} \right)}}dx} = x + \cos x + c$.

Therefore, the correct option is B.

Note: For such a question, the best way is to write the given expression in terms of $\sin x$ and $\cos x$. Also, we need to be thorough with the trigonometric properties as well as integration formulas. While integrating, we forget to add the constant term and usually ignore the constant term. So, this needs to be taken care of. Also, while integrating $\sin x$ and $\cos x$, we usually get confused about where a negative sign is included and where it is not included.