Question

Find the integral of the function $\dfrac{1-\cos x}{1+\cos x}$

Answer 1

Hint : Remember the Trigonometric Identities ! There’s always a hint in the question if you observe it carefully.
Trigonometric identities that might be helpful are $\cos 2\theta =2{{\cos }^{2}}\theta -1$ , $\cos 2\theta =1-2{{\sin }^{2}}\theta$ etc.

Complete step-by-step answer :
Trigonometric identities used are:
$\cos 2\theta =2{{\cos }^{2}}\theta -1$ ,
$\cos 2\theta =1-2{{\sin }^{2}}\theta$ ,
${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$ .
We need to find $\int{\dfrac{1-\cos x}{1+\cos x}}$ .
We know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$ (trigonometric identity)
$\cos 2\theta +1=2{{\cos }^{2}}\theta$
${{\cos }^{2}}\theta =\dfrac{\cos 2\theta +1}{2}$
Replacing $\theta$ with $\dfrac{x}{2}$ , we get
${{\cos }^{2}}\dfrac{x}{2}=\dfrac{\cos 2\dfrac{x}{2}+1}{2}$
${{\cos }^{2}}\dfrac{x}{2}=\dfrac{\cos x+1}{2}$
$2{{\cos }^{2}}\dfrac{x}{2}=\cos x+1$ --equation 1
Similarly, we know that $\cos 2\theta =1-2{{\sin }^{2}}\theta$ (trigonometric identity)
$2{{\sin }^{2}}\theta =1-\cos 2\theta$
Replacing $\theta$ with $\dfrac{x}{2}$ , we get
$2{{\sin }^{2}}\dfrac{x}{2}=1-\cos x$ --equation 2
Substituting equation 1 and equation 2 , we get
$\Rightarrow \int{\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}}$
$\Rightarrow \int{{{\tan }^{2}}\dfrac{x}{2}}$
We know that ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$ (trigonometric identity)
Replacing $\theta$ with $\dfrac{x}{2}$ , we get
$\Rightarrow {{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}-1$
Therefore,
$\Rightarrow \int{\left( {{\sec }^{2}}\dfrac{x}{2}-1 \right)}dx$
$\Rightarrow \int{{{\sec }^{2}}\dfrac{x}{2}}dx$ $-\int{dx}$ --equation 3
Let’s find $\int{{{\sec }^{2}}\dfrac{x}{2}}dx$ by substitution method:
Let $\dfrac{x}{2}$ be u, then on differentiating both sides we get, $\dfrac{1}{2}dx=du$
$dx=2du$
$\Rightarrow \int{{{\sec }^{2}}\dfrac{x}{2}}dx=2\int{{{\sec }^{2}}u}du$
We know that $\int{{{\sec }^{2}}xdx=\tan x}$ +C (trigonometric identity)
Where C is the integral constant
Therefore, $2\int{{{\sec }^{2}}u}du$ = $2\left( \tan u+C \right)$
Replacing u with $\dfrac{x}{2}$ , we get
$\Rightarrow 2\tan \dfrac{x}{2}+C$ --equation 4
(2C is also a constant represented by C)
We know that, $\int{dx}$ = x+C --equation 5
(C being the integral constant)
Substituting equation 4 and equation 5 in equation 3, we get
$\Rightarrow 2\tan \dfrac{x}{2}-x+C$
Hence, $\int{\dfrac{1-\cos x}{1+\cos x}}$ = $2\tan \dfrac{x}{2}-x+C$
So, the correct answer is “ $\int{\dfrac{1-\cos x}{1+\cos x}}$ = $2\tan \dfrac{x}{2}-x+C$ ”.

Note : Keeping in mind some standard integral rules can save time and solve the question much faster. For example, in this question one need not show integration of ${{\sec }^{2}}x$ and can directly solve the question. This will take less time to solve the question. Similarly one can remember other standard integral rules, as many as possible!