Integrals of Other Trig Functions: Videos & Practice Problems

Example 1: Integral of Tangent & Cotangent

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:

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

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

Example 2: Integral of Secant & Cosecant

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:

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

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