Question

How do you find the antiderivative of $\dfrac{1}{1-\cos x}$?

Answer 1

Rationalize the denominator by multiplying and dividing the given function with $\left( 1+\cos x \right)$. Use the identity: - ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ to simplify the denominator. Now, break the terms and write the function as: - $\int{{{\csc }^{2}}xdx}+\int{\csc x\cot xdx}$. Now, use the formulas: - $\int{\csc x\cot xdx}=-\csc x$ and $\int{{{\csc }^{2}}xdx}=-\cot x$ to get the answer. Add the constant of indefinite integration ‘c’ to complete the integral.

Complete step by step solution:
Here, we have been provided with the function $\dfrac{1}{1-\cos x}$ and we are asked to determine its anti-derivative which means integral. Now, let us assume the required integral as ‘I’. So, we have,
$\Rightarrow I=\int{\dfrac{1}{1-cosx}dx}$
Now, rationalizing the denominator by multiplying and dividing the function with $\left( 1+\cos x \right)$, we get,
$\Rightarrow I=\int{\dfrac{\left( 1+\cos x \right)}{\left( 1-\cos x \right)\left( 1+\cos x \right)}dx}$
Using the algebraic identity: - $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get,
$\Rightarrow I=\int{\left[ \dfrac{1+\cos x}{\left( 1-{{\cos }^{2}}x \right)} \right]dx}$
We know that, ${{\cos }^{2}}x+{{\sin }^{2}}x=1$, so we have $\left( 1-{{\cos }^{2}}x \right)={{\sin }^{2}}x$. Therefore, the integral ‘I’ becomes: -
$\Rightarrow I=\int{\left( \dfrac{1+\cos x}{{{\sin }^{2}}x} \right)dx}$
Breaking the integral terms, we get,
$\Rightarrow I=\int{\dfrac{1}{{{\sin }^{2}}x}dx}+\int{\dfrac{\cos x}{{{\sin }^{2}}x}dx}$
The above expression can be simplified as: -
$\Rightarrow I=\int{{{\csc }^{2}}xdx}+\int{cscx\cot xdx}$
Using the basic formulas: - $\int{\csc x\cot xdx}=-\csc x$ and $\int{{{\csc }^{2}}xdx}=-\cot x$, we get,

& \Rightarrow I=-\cot x-\csc x \\\ & \Rightarrow I=-\left( \cot x+\csc x \right) \\\ \end{aligned}$$ Since, the required integral was an indefinite integral, so we add a constant of indefinite integration ‘c’ at the end. So, we have, $$\Rightarrow I=-\left( \cot x+\csc x \right)+c$$, where ‘c’ = constant of integration. **Note:** One may note that we can also solve the question by using a different approach. What we can do is we will apply the half angle formula given as: - $$\cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$$. Then in the numerator we will use the identity $$1+{{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}$$ for the simplification. In the next step we will substitute $$\tan \left( \dfrac{x}{2} \right)=k$$ and $${{\sec }^{2}}\dfrac{x}{2}dx=2dk$$. In this case the obtained integral will be $$\left( -\cot \dfrac{x}{2} \right)$$. Actually, when you will simplify $$-\left( \cot x+\csc x \right)$$ using the formulas: - $$1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$$ and $$\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$$, you will get $$\left( -\cot \dfrac{x}{2} \right)$$, so you can conclude that the answer we have obtained is correct and only the form is different. Do not forget to add the constant of indefinite integration ‘c’ in the end otherwise the answer would be considered incomplete.