Question

Show that the two formulas are equivalent. $$ \begin{array}{l} \int \tan x d x=-\ln |\cos x|+C \\ \int \tan x d x=\ln |\sec x|+C \end{array} $$

Short answer

The two integral formulas for \(\tan x\) are shown to be equivalent because \(\sec x = \frac{1}{\cos x}\), and with the application of relevant logarithm properties, we arrive at the same formula.

Step-by-step solution

Understand the properties of trigonometric functions

It is known from basic trigonometric identities that \(\cos x = \frac{1}{\sec x}\).The absolute value function in logarithms can be ignored when proving equivalence as the individual identities are equivalent.

Substitute the trigonometric identity into the second formula

Substitute \(\sec x= \frac{1}{\cos x}\) into the second formula, thus \(\ln |\sec x|= \ln |\frac{1}{\cos x}|\).

Use the logarithmic property

Using the logarithmic property of \(\ln(a/b) = \ln(a) - \ln(b)\) and \(\ln(1/a)=-\ln(a)\), we can convert \(\ln |\frac{1}{\cos x}|\) into \(- \ln |\cos x|\).

Comparison

Compare the formula derived with the first formula. They are both \(-\ln |\cos x|+C\), meaning they are equivalent.