Question

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

Answer 1

Hint: In this problem they have asked to calculate the antiderivative of the given equation. We know that antiderivative is nothing but integration. So first we need to simplify the given equation to integrate. In the problem we have the function $\dfrac{1}{\cos x-1}$. We can’t integrate the given equation directly, so first we need to simplify the equation. For this we will use the conjugate multiplication trick which is nothing but multiplying and dividing the given equation with conjugate of $\cos x-1$ which is $\cos x+1$. So, we will divide and multiply the given equation with $\cos x+1$ and use the trigonometric formula and simplify the equation. Once we have the simplified form of the given equation, we will integrate the equation and use integration formulas to get the final result.

Complete step-by-step solution:
Given equation $\dfrac{dx}{\cos x-1}$.
Considering the equation $\dfrac{1}{\cos x-1}$ separately.
For the above equation we can’t do integration directly. So, we need to simplify the above equation. For this we are going to multiply and divide the above equation with $\cos x+1$, then we will get
$\dfrac{1}{\cos x-1}=\dfrac{1}{\cos x-1}\times \dfrac{\cos x+1}{\cos x+1}$
We have the algebraic formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. Applying this formula in the above equation, then we will get
$\Rightarrow \dfrac{1}{\cos x-1}=\dfrac{\cos x+1}{{{\cos }^{2}}x-1}$
We have the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. From this identity we are going to substitute the value of ${{\cos }^{2}}x-1=-{{\sin }^{2}}x$ in the above equation and simplifying the equation, then we will get
$\begin{align}
  & \Rightarrow \dfrac{1}{\cos x-1}=\dfrac{\cos x+1}{-{{\sin }^{2}}x} \\
 & \Rightarrow \dfrac{1}{\cos x-1}=-\dfrac{\cos x}{{{\sin }^{2}}x}-\dfrac{1}{{{\sin }^{2}}x} \\
\end{align}$
We know that $\dfrac{\cos x}{\sin x}=\cot x$, $\dfrac{1}{{{\sin }^{2}}x}={{\csc }^{2}}x$, then we will get
$\Rightarrow \dfrac{1}{\cos x-1}=-\cot x.\csc x-{{\csc }^{2}}x$
Now integrating the given equation by using the value of $\dfrac{1}{\cos x-1}$, then we will get
$\int{\dfrac{dx}{\cos x-1}=\int{-\cot x.\csc xdx}+\int{-{{\csc }^{2}}xdx}}$
We know that $\int{-\csc x.\cot xdx}=\csc x+C$, $\int{-{{\csc }^{2}}xdx}=\cot x+C$. Substituting these values in the above equation, then we will get
$\therefore \int{\dfrac{dx}{\cos x-1}}=\csc x+\cot x+C$

Note: In this problem we have calculated the antiderivative of $\dfrac{dx}{\cos x-1}$. We can also calculate the antiderivative of $\dfrac{dx}{\cos x+1}$ by using the same method. But the difference is we have multiplied and divided $\dfrac{1}{\cos x-1}$ with $\cos x+1$. But for $\dfrac{dx}{\cos x+1}$ we will multiply and divide with $\cos x-1$.