Cotangent Integral Formula

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

The integration of 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 formulas \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 derivative, so the integral sign \int {} and \frac{d}{{dx}} on the right side will cancel each other, i.e.

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

Alternate Proof:

We have given integration of the form

\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 Formulas of Cotangent Function:
The other formulas of cotangent integral with angle of sine is in the form of 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