# Lesson Explainer: Indefinite Integrals: Reciprocal Trigonometric Functions Mathematics

In this explainer, we will learn how to find indefinite integrals of functions that result in reciprocal trigonometric functions.

We can identify a variety of indefinite integrals involving reciprocal trigonometric functions through the associated derivatives. We first recall the derivative: This leads to the following indefinite integral:

### Standard Result: Indefinite Integral of the Product of Secant and Tangent Functions

In our first example, we will use this formula to solve an indefinite integral.

Determine .

### Answer

We note that the given integrand contains the produce of secant and tangent. So, we recall the following indefinite integral:

To use this formula for our example, we need to modify the argument of the trigonometric functions. We can do this by using the -substitution

Substituting this change of variables to our integral, we have

We can now apply our formula we recalled earlier to write this indefinite integral as

Since is an arbitrary constant, we can simply denote the constant as in our answer. Substituting back into the resulting expression, we obtain

In our next example, we will solve an indefinite integral which requires us to simplify the integrand first.

Determine .

### Answer

When we have a factored expression within the integrand, as is the case here, we should start by expanding through the parenthesis in the integrand. We obtain

Since , we can simplify the first term using the fact that

Substituting this into our integrand, we obtain

Now, we need to evaluate the following indefinite integral:

The first term in the integrand involves a cosine, and the second term involves the product of a secant and a tangent. We recall the following formulas:

Before we can apply these formulas to solve the given indefinite integral, we need to modify the argument of the trigonometric functions. We use the -substitution

We can rearrange the second equation to state . Substituting this change of variable into our integral, we get

We apply the formulas above to obtain for some arbitrary constants and . Since, after distributing through the parenthesis, we will end up with a combination of and , we can replace this expression by another arbitrary constant to write the solution as

Substituting back into the expression, we obtain the indefinite integral

Let us consider another example involving the product of and .

Determine .

### Answer

Since there is a factored expression within the integrand, we should start by expanding through the parenthesis in the integrand. We obtain

The first term is the product of the secant and tangent function, and the second term is the square of the secant function. In order to solve our problem, we recall the following indefinite integrals:

Applying these formulae to our integral, we obtain for some arbitrary constants and . Since, after distributing through the parenthesis, we will end up with a combination of and , we can replace this expression by another arbitrary constant to write the solution as

In the previous examples, we used the formula for the indefinite integral that have results in the form of the secant and tangent functions. We now consider integrals that involve the complementary trigonometric functions and . Recall that

Hence, integrating both sides of the equation, we obtain the following indefinite integral.

### Standard Result: Indefinite Integral of the Product of Cosecant and Cotangent Functions

To help with remembering this formula, we note the resemblance between this formula and the first formula we stated, which is

From this integral, we can replace each trigonometric function with their complements and place a negative sign on the function on the right-hand side of the equation to retrieve the new formula. This resemblance is not coincidental. Let us examine why this is expected in general.

We recall that the complementary trigonometric functions cosine, cotangent, and cosecant take the complementary angles of their counterparts. In other words,

Therefore, we can write

We can use the -substitution method by defining

We can write the last equation as . Substituting this change of variable into our indefinite integral, we get

Applying the formula for the antiderivative of , we obtain

Replacing back into the resulting expression, we obtain

We note that, when integrating complementary trigonometric functions, we expect a negative sign resulting from the substituting , which leads to . Other than the negative sign, we can retrieve the formula for the integral of complementary trigonometric functions by their original counterparts.

Let us evaluate an indefinite integral using this formula.

Determine .

### Answer

The given integrand is the product of the cosecant and cotangent functions. Hence, we recall the following integral:

Before applying this formula to our integral, let us use the -substitution method by defining

We can also write the last equation as . Substituting this change of variables to our integral, we have

We can now apply our formula we recalled earlier to write this indefinite integral as

Since is an arbitrary constant, we can simply denote the constant as in our answer. Substituting back into the resulting expression, we obtain

For our final formula, we state the indefinite integral of . To obtain this integral, we first recall the integral of :

As discussed earlier, we can obtain the complementary counterpart of this integral by replacing and by their complements and and placing a negative sign on the right-hand side.

### Standard Result: Indefinite Integral of the Square of the Cosecant Function

When integrating or , we can use the following trigonometric identity to express the function in terms of or respectively:

Both of these identities can be derived from the Pythagorean identity when both sides of the equation are divided by either or .

In our final example, we will evaluate an indefinite integral involving by, first, applying this trigonometric identity and, second, applying the formula for the antiderivative of .

Determine .

### Answer

Since is a constant, we can begin by bringing it outside the integral:

We can use the -substitution method by defining

We can also write the last equation as . Substituting this change of variables to our integral, we obtain

We note that the integrand contains the square of cotangent function. The antiderivative of is not readily available, but we know that can be expressed in terms of and we know the integral of :

To obtain the relation between and , we divide both sides of the Pythagorean identity by :

This is the same as

We note that our integrand is precisely . Hence, applying this identity to our integral, we get

We can now apply the formula for the antiderivative of to evaluate this indefinite integral:

We can replace by since is an arbitrary constant. We can also replace by . Hence,

Let us recap a few important concepts from this explainer.

### Key Points

• The indefinite integral of the product of and is
• Indefinite integrals involving the complementary trigonometric functions can be obtained by replacing each trigonometric function with its complementary counterpart and placing a negative sign on the result of the integral. The important indefinite integrals involving complementary reciprocal trigonometric functions are
• Integrals of or are not readily available. To compute an integral involving either of these functions, we can use the following trigonometric identities to relate it to or respectively:

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