Question

Evaluate the following integral $\int{\dfrac{1-\cos x}{1+\cos x}dx}$

Answer 1

First of all we will find the value of $\cos x$ by converting it into $\cos \left( \dfrac{x}{2}+\dfrac{x}{2} \right)$ and then we will use the formula $\cos \left( A+B \right)=\cos A.\cos B-\operatorname{Sin}A.\sin B$ to find the value of $\cos x$. Now we will calculate the values of $1-\cos x$ and $1+\cos x$ individually by using the value $\cos x$ obtained above. Here we will use the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to find the values of $1-\cos x$ and $1+\cos x$. From the values of $1-\cos x$ and $1+\cos x$ we will calculate the value of $\dfrac{1-\cos x}{1+\cos x}$, and then we will integrate the obtained value to get the result.

Complete step-by-step solution:
Given that, $\int{\dfrac{1-\cos x}{1+\cos x}dx}$
We are going to write the $x$ value as $\dfrac{x}{2}+\dfrac{x}{2}$ in $\cos$function to find the value of $\cos x$, then we will get
$\cos x=\cos \left( \dfrac{x}{2}+\dfrac{x}{2} \right)$
We know the formula $\cos \left( A+B \right)=\cos A.\cos B-\operatorname{Sin}A.\sin B$, then the value of $\cos \left( \dfrac{x}{2}+\dfrac{x}{2} \right)$ will be written as
$\begin{aligned} & \cos x=\cos \left( \dfrac{x}{2}+\dfrac{x}{2} \right) \\\ & =\cos \dfrac{x}{2}.\cos \dfrac{x}{2}-\sin \dfrac{x}{2}.\sin \dfrac{x}{2} \\\ \end{aligned}$
We know that $a.a={{a}^{2}}$, hence the values of $\cos \dfrac{x}{2}.\cos \dfrac{x}{2}$ and $\sin \dfrac{x}{2}.\sin \dfrac{x}{2}$ will be written as
$\begin{aligned} & \cos x=\cos \dfrac{x}{2}.\cos \dfrac{x}{2}-\sin \dfrac{x}{2}.\sin \dfrac{x}{2} \\\ & ={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \\\ \end{aligned}$
Here we got the value of $\cos x$ as ${{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}$, i.e.
$\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}....\left( \text{i} \right)$
Now we are going to calculate the values of $1-\cos x$ and $1+\cos x$ from the above equation.
We will add $1$ on both sides of equation $\left( \text{i} \right)$ to get the value of $1+\cos x$, then
$\begin{aligned} & \cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \\\ &\Rightarrow 1+\cos x=1+{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \\\ \end{aligned}$
Rearranging the term in Left Hand Side as
$1+\cos x={{\cos }^{2}}\dfrac{x}{2}+\left( 1-{{\sin }^{2}}\dfrac{x}{2} \right)$
We have trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. From this identity we can get the value$1-{{\sin }^{2}}A={{\cos }^{2}}A$. Then we will have
$\begin{aligned} & 1+\cos x={{\cos }^{2}}\dfrac{x}{2}+\left( 1-{{\sin }^{2}}\dfrac{x}{2} \right) \\\ & ={{\cos }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2} \\\ \end{aligned}$
We know that $a+a=2a$, then we will get
$\begin{aligned} & 1+\cos x={{\cos }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2} \\\ &\Rightarrow 1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}....\left( \text{ii} \right) \\\ \end{aligned}$
Now we are going to subtract the value of $\cos x$ obtained in equation $\left( \text{i} \right)$ from $1$ to get the value of $1-\cos x$, then we will have
$1-\cos x=1-\left( {{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \right)$
When we multiplied a negative sign/integer with positive sign/integer we will get negative sign/integer at the same time we will get positive sign/integer when we multiplied a negative sign/integer with the negative sign/integer, then
$\begin{aligned} & 1-\cos x=1-\left( {{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \right) \\\ & =1-{{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2} \\\ \end{aligned}$
We have trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. From this identity we can get the value$1-{{\cos }^{2}}A={{\sin }^{2}}A$. Then we will have
$\begin{aligned} & 1-\cos x=1-{{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2} \\\ &\Rightarrow 1-\cos x={{\sin }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2} \\\ &\Rightarrow 1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}.....\left( \text{iii} \right) \\\ \end{aligned}$
From the equations $\left( \text{ii} \right)$ and $\left( \text{iii} \right)$ the value of $\dfrac{1-\cos x}{1+\cos x}$ is
$\begin{aligned} & \dfrac{1-\cos x}{1+\cos x}=\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \\\ & ={{\tan }^{2}}\dfrac{x}{2} \\\ \end{aligned}$
We have another trigonometric identity ${{\sec }^{2}}A-{{\tan }^{2}}A=1$. From this identity we have the value ${{\tan }^{2}}A={{\sec }^{2}}A-1$, then
$\begin{aligned} & \dfrac{1-\cos x}{1+\cos x}={{\tan }^{2}}\dfrac{x}{2} \\\ & ={{\sec }^{2}}\dfrac{x}{2}-1 \\\ \end{aligned}$
Now integrating the above equation, then
$\begin{aligned} & \int{\dfrac{1-\cos x}{1+\cos x}dx}=\int{\left( {{\sec }^{2}}\dfrac{x}{2}-1 \right)}dx \\\ & =\int{{{\sec }^{2}}\dfrac{x}{2}dx}-\int{dx} \\\ \end{aligned}$
We have $\int{{{\sec }^{2}}A}dA=\tan A+C$ and $\int{1dx=x+C}$, substituting above value in the integration, then
$\int{\dfrac{1-\cos x}{1+\cos x}dx}=2\tan \dfrac{x}{2}-x+C$
Hence, we have $\int{\dfrac{1-\cos x}{1+\cos x}dx}=2\tan \dfrac{x}{2}-x+C$

Note: In this problem we do many substitutions from trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$, while substituting value there is a chance of making a mistake when you don’t write this identity in the problem. It is advisable to write the trigonometric identities when we are going to use that identity.