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. cos(x) + C
• D. ln|tan(x)| + C

Answer: (A)

The integral of \( \tan(x) \, dx \) can be derived using the identity for tangent: \[ \tan(x) = \frac{\sin(x)}{\cos(x)} \] This allows us to rewrite the integral as: \[ \int \tan(x) \, dx = \int \frac{\sin(x)}{\cos(x)} \, dx \] To solve this integral, we can use the substitution method. Let \( u = \cos(x) \), which implies that \( du = -\sin(x) \, dx \) or \( -du = \sin(x) \, dx \). The integral then transforms to: \[ \int \frac{\sin(x)}{\cos(x)} \, dx = -\int \frac{1}{u} \, du \] This integral evaluates to: \[ -\ln|u| + C = -\ln|\cos(x)| + C \] Thus, we arrive at the integral of \( \tan(x) \, dx \): \[ -\ln|\cos(x)| + C \] This derivation confirms that the correct answer aligns with the option that states \( -\ln|

The integral of ( \tan(x) , dx ) can be derived using the identity for tangent:

[

\tan(x) = \frac{\sin(x)}{\cos(x)}

]

This allows us to rewrite the integral as:

[

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

]

To solve this integral, we can use the substitution method. Let ( u = \cos(x) ), which implies that ( du = -\sin(x) , dx ) or ( -du = \sin(x) , dx ). The integral then transforms to:

[

\int \frac{\sin(x)}{\cos(x)} , dx = -\int \frac{1}{u} , du

]

This integral evaluates to:

[

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

]

Thus, we arrive at the integral of ( \tan(x) , dx ):

[

-\ln|\cos(x)| + C

]

This derivation confirms that the correct answer aligns with the option that states ( -\ln|