Cotangent Integral Formula

In this tutorial we will prove the formula of cotangent integral which is another important formula in integral calculus. This integral belongs to the trigonometric formulae.

The integration of the cotangent function is of the form

\int {\cot xdx = } \ln \sin x + c

To prove this formula, consider

\frac{d}{{dx}}\left[ {\ln \sin x + c} \right] = \frac{d}{{dx}}\ln \sin x + \frac{d}{{dx}}c

Using the derivative formula \frac{d}{{dx}}\ln x = \frac{1}{x} and \frac{d}{{dx}}\sin x = \cos x, we have

\begin{gathered} \frac{d}{{dx}}\left[ {\ln \sin x + c} \right] = \frac{1}{{\sin x}}\frac{d}{{dx}}\sin x + 0 \\ \Rightarrow \frac{d}{{dx}}\left[ {\ln \sin x + c} \right] = \frac{1}{{\sin x}}\cos x \\ \Rightarrow \frac{d}{{dx}}\left[ {\ln \sin x + c} \right] = \cot x \\ \Rightarrow \cot x = \frac{d}{{dx}}\left[ {\ln \sin x + c} \right] \\ \Rightarrow \cot xdx = d\left[ {\ln \sin x + c} \right]\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}

Integrating both sides of equation (i) with respect to x, we have

\int {\cot xdx} = \int {d\left[ {\ln \sin x + c} \right]}

As we know that by definition integration is the inverse process of the derivative, the integral sign \int {} and \frac{d}{{dx}} on the right side will cancel each other out, i.e.

\int {\cot xdx = } \ln \sin x + c

Alternate Proof

We substitute \cot x  with \frac{{\cos x}}{{\sin x}}

\int {\cot xdx = \int {\frac{{\cos x}}{{\sin x}}dx} }

Here we have f\left( x \right) = \sin x then f'\left( x \right) = - \cos x

\int {\cot xdx = \int {\frac{{\cos x}}{{\sin x}}dx} }

Using the formula of integration, \int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \ln f\left( x \right) + c}

\int {\cot xdx = \ln \sin x + c}

Other Integral Formulae of the Cotangent Function
The other formulae of cotangent integral with an angle of sine in the form of a function are given as

1. \int {\cot axdx = \frac{1}{a}\ln \sin ax + c}

2. \int {\cot f\left( x \right)f'\left( x \right)dx = \ln \sin f\left( x \right) + c}

Example: Evaluate the integral \int {\cot 3xdx} with respect to x

We have integral

I = \int {\cot 3xdx}

Using the formula \int {\cot axdx = \frac{1}{a}\ln \sin ax + c} , we have

\int {\cot 3xdx} = \frac{1}{3}\ln \sin 3x + c