What is the integral of tan(x) dx?

What is the integral of tan(x) dx?

• A. -ln|cos(x)| + C
• B. ln|sin(x)| + C
• C. ln|tan(x)| + C
• D. -ln|sin(x)| + C

Answer: (A)

To find the integral of \(\tan(x) dx\), we can use the relationship between tangent and sine and cosine. Recall that \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). To integrate \(\tan(x)\), we can rewrite it as: \[ \int \tan(x) dx = \int \frac{\sin(x)}{\cos(x)} dx. \] A common method to integrate this function is to use substitution. Let \(u = \cos(x)\). Then, the derivative of \(u\) with respect to \(x\) is \(-\sin(x) dx\), which implies that \(dx = -\frac{du}{\sin(x)}\). Substituting these into the integral, we can write: \[ \int \tan(x) dx = -\int \frac{1}{u} du. \] The integral of \(\frac{1}{u}\) is \(\ln|u| + C\), leading us to: \[ -\ln|u| + C = -\ln|\cos(x)| + C. \] Thus, the integral of \(\tan(x) dx\)

To find the integral of (\tan(x) dx), we can use the relationship between tangent and sine and cosine. Recall that (\tan(x) = \frac{\sin(x)}{\cos(x)}).

To integrate (\tan(x)), we can rewrite it as:

[

\int \tan(x) dx = \int \frac{\sin(x)}{\cos(x)} dx.

]

A common method to integrate this function is to use substitution. Let (u = \cos(x)). Then, the derivative of (u) with respect to (x) is (-\sin(x) dx), which implies that (dx = -\frac{du}{\sin(x)}).

Substituting these into the integral, we can write:

[

\int \tan(x) dx = -\int \frac{1}{u} du.

]

The integral of (\frac{1}{u}) is (\ln|u| + C), leading us to:

[

-\ln|u| + C = -\ln|\cos(x)| + C.

]

Thus, the integral of (\tan(x) dx)