Integrals of Other Trig Functions: Videos & Practice Problems

To find the integral of tangent \( x \), we start by rewriting it in terms of sine and cosine. The integral can be expressed as:

\[ \int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx \]

Next, we use substitution. Let \( u = \cos x \), which gives us \( du = -\sin x \, dx \). Rearranging this, we have \( -du = \sin x \, dx \). Substituting these into the integral, we get:

\[ \int \tan x \, dx = -\int \frac{1}{u} \, du \]

This integral is straightforward to evaluate, yielding:

\[ -\ln |u| + C \]

Substituting back for \( u \), we find:

\[ -\ln |\cos x| + C \]

Using properties of logarithms, we can rewrite this as:

\[ \ln |\sec x| + C \]

Thus, the integral of tangent \( x \) is:

\[ \int \tan x \, dx = \ln |\sec x| + C \]

Now, let's consider the integral of cotangent \( x \). We rewrite cotangent in terms of sine and cosine:

\[ \int \cot x \, dx = \int \frac{\cos x}{\sin x} \, dx \]

Using a similar substitution, let \( u = \sin x \), which gives \( du = \cos x \, dx \). This allows us to rewrite the integral as:

\[ \int \cot x \, dx = \int \frac{1}{u} \, du \]

Evaluating this integral results in:

\[ \ln |u| + C \]

Substituting back for \( u \), we have:

\[ \ln |\sin x| + C \]

Therefore, the integral of cotangent \( x \) is:

\[ \int \cot x \, dx = \ln |\sin x| + C \]

In summary, the integrals of the tangent and cotangent functions are:

\[ \int \tan x \, dx = \ln |\sec x| + C \]

\[ \int \cot x \, dx = \ln |\sin x| + C \]

To find the integral of secant and cosecant functions, we can utilize strategic substitutions and properties of logarithms. Starting with the integral of secant, we consider the integral:

\[\int \sec x \, dx\]

To simplify this, we multiply by the fraction \[\frac{\sec x + \tan x}{\sec x + \tan x}\], which is effectively multiplying by one. This gives us:

\[\int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx\]

Next, we set \[u = \sec x + \tan x\]. The derivative \[du\] is calculated as:

\[du = (\sec x \tan x + \sec^2 x) \, dx\]

Recognizing that the integrand matches \[du\], we can rewrite the integral as:

\[\int \frac{1}{u} \, du\]

This integral evaluates to:

\[\ln |u| + C = \ln |\sec x + \tan x| + C\]

Thus, the integral of secant is:

\[\int \sec x \, dx = \ln |\sec x + \tan x| + C\]

Next, we turn to the integral of cosecant:

\[\int \csc x \, dx\]

Similar to the previous case, we multiply by the fraction \[\frac{\csc x + \cot x}{\csc x + \cot x}\], leading to:

\[\int \frac{\csc^2 x + \csc x \cot x}{\csc x + \cot x} \, dx\]

Setting \[u = \csc x + \cot x\], we find the derivative:

\[du = (-\csc x \cot x - \csc^2 x) \, dx\]

To match the integrand with \[du\], we multiply the entire integral by -1, resulting in:

\[-\int \frac{1}{u} \, du\]

This evaluates to:

\[-\ln |u| + C = -\ln |\csc x + \cot x| + C\]

Thus, the integral of cosecant is:

\[\int \csc x \, dx = -\ln |\csc x + \cot x| + C\]

In summary, the integral rules for these trigonometric functions are:

\[\int \sec x \, dx = \ln |\sec x + \tan x| + C\]

\[\int \csc x \, dx = -\ln |\csc x + \cot x| + C\]