Question

How do you find the integral of $\int {\dfrac{1}{{1 + \cos (x)}}} ?$

Answer 1

Hint : The question describes the operation of addition/ subtraction/ multiplication/ division. We need to know algebraic formulae to simplify the term in the integral. Also, we need to know trigonometrical conditions and basic integration formulae. We need to know how to substitute the trigonometric identities at the correct place to solve the given problem. Also, we need to know how to take conjugate to simplify the denominator.

Complete step-by-step answer :
In this question, we have to find the integral form of $\dfrac{1}{{1 + \cos x}}$ . That is given below,
$\int {\dfrac{1}{{1 + \cos (x)}}} dx = ?$
For solving the above equation we take conjugate for the term $\dfrac{1}{{1 + \cos x}}$ . So, we get

$$\int {\dfrac{1}{{1 + \cos x}}} dx = \int {\dfrac{1}{{\left( {1 + \cos x} \right)}}} \times \dfrac{{\left( {1 - \cos x} \right)}}{{\left( {1 - \cos x} \right)}}dx \\\ \int {\dfrac{1}{{1 + \cos x}}} dx = \int {\dfrac{{1 - \cos x}}{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}} dx \to \left( 1 \right) \\\$$

We know that,
$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} \to \left( 2 \right)$
Let’s substitute the equation $\left( 2 \right)$ in the equation $\left( 1 \right)$ , we get
$\int {\dfrac{{1 - \cos x}}{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}} dx = \int {\dfrac{{1 - \cos x}}{{\left( {1 - {{\cos }^2}x} \right)}}} dx \to \left( 3 \right)$
We know that,
${\sin ^2}x + {\cos ^2}x = 1$
The above equation can also be written as,
${\sin ^2}x = 1 - {\cos ^2}x \to \left( 4 \right)$
Let’s substitute the above equation in the equation $\left( 3 \right)$ , we get
$\int {\dfrac{{1 - \cos x}}{{\left( {1 - {{\cos }^2}x} \right)}}} dx = \int {\dfrac{{1 - \cos x}}{{{{\sin }^2}x}}} dx$
The above equation can also be written as,
$\int {\dfrac{{1 - \cos x}}{{{{\sin }^2}x}}} dx = \int {\dfrac{1}{{{{\sin }^2}x}}} dx - \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}} dx \to \left( 5 \right)$
We know that,
$\dfrac{1}{{\sin x}} = \cos ecx$
So,
$\dfrac{1}{{{{\sin }^2}x}} = \cos e{c^2}x \to \left( 6 \right)$
Let’s substitute the above equation in the equation $\left( 5 \right)$ , we get
$\int {\dfrac{1}{{{{\sin }^2}x}}} dx - \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}} dx = \int {\cos ec(x)dx} - \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}} dx \to \left( 7 \right)$
First, we have to solve,
$\int {\dfrac{{\cos x}}{{{{\sin }^2}x}}} dx = ?$
Here we take,

$$u = \sin x \\\ du = \cos xdx \;$$

So, we get
$\int {\dfrac{{\cos x}}{{{{\sin }^2}x}}} dx = \int {\dfrac{1}{{{u^2}}}} du$
So, the equation $\left( 7 \right)$ becomes,
$\int {\cos ec(x)dx} - \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}} dx = \int {\cos ec(x)dx - \int {\dfrac{1}{{{u^2}}}} } du \to \left( 8 \right)$
We know that,
$\int {\cos ec(x)dx = - \cot x}$
And
$\int {\dfrac{1}{{{u^2}}}} du = \int {{u^{ - 2}}du} = \dfrac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}} = \dfrac{{{u^{ - 1}}}}{{ - 1}} = \dfrac{{ - 1}}{u}$
We know that
$u = \sin x$
So we get,
$\int {\dfrac{1}{{{u^2}}}} du = \dfrac{{ - 1}}{u} = \dfrac{{ - 1}}{{\sin x}}$
So, the equation $\left( 8 \right)$ becomes,
$\int {\cos e{c^2}xdx - \int {\dfrac{1}{{{u^2}}}} } du = - \cot x + \dfrac{1}{{\sin x}} + c$
The above equation can also be written as,
$\int {\cos e{c^2}xdx - \int {\dfrac{1}{{{u^2}}}} } du = - \cot x + \cos ec(x) + c$
So, the final answer is,
$\int {\cos e{c^2}xdx - \int {\dfrac{1}{{{u^2}}}} } du = - \cot x + \cos ec(x) + c$
So, the correct answer is “$- \cot x + \cos ec(x) + c$”.

Note : When we replace $u$ with $\sin x$ we should add $c$ in the final answer. If we want to simplify the denominator term we can take conjugate for that term and multiply it with both numerator and denominator. To solve these types of questions we would remember the basic formulae for integration and trigonometric identities