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: ddsecsectan𝑥𝑥=𝑥𝑥. This leads to the following indefinite integral:

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

𝑥𝑥𝑥=𝑥+sectandsecC

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

Example 1: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions

Determine (93𝑥3𝑥)𝑥tansecd.

Answer

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

To use this formula for our example, we need to modify the argument 3𝑥 of the trigonometric functions. We can do this by using the 𝑢-substitution 𝑢=3𝑥,𝑢=3𝑥.whichimpliesdd

Substituting this change of variables to our integral, we have (93𝑥3𝑥)𝑥=(33𝑥3𝑥)3𝑥=3𝑢𝑢𝑢.tansecdtansecdtansecd

We can now apply our formula we recalled earlier to write this indefinite integral as 3(𝑢+)=3𝑢+3.secCsecC

Since C is an arbitrary constant, we can simply denote the constant 3C as C in our answer. Substituting 𝑢=3𝑥 back into the resulting expression, we obtain (93𝑥3𝑥)𝑥=33𝑥+.tansecdsecC

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

Example 2: Finding the Integration of a Function Involving Trigonometric Functions

Determine 87𝑥47𝑥+67𝑥𝑥seccostand.

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 87𝑥47𝑥+67𝑥=327𝑥7𝑥487𝑥7𝑥.seccostanseccossectan

Since seccos7𝑥=17𝑥, we can simplify the first term using the fact that seccoscoscoscos7𝑥7𝑥=7𝑥7𝑥=7𝑥.

Substituting this into our integrand, we obtain 327𝑥7𝑥487𝑥7𝑥=327𝑥487𝑥7𝑥.seccossectancossectan

Now, we need to evaluate the following indefinite integral: (327𝑥487𝑥7𝑥)𝑥.cossectand

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: 𝑥𝑥=𝑥+,𝑥𝑥𝑥=𝑥+.cosdsinCsectandsecC

Before we can apply these formulas to solve the given indefinite integral, we need to modify the argument 7𝑥 of the trigonometric functions. We use the 𝑢-substitution 𝑢=7𝑥,𝑢=7𝑥.whichimpliesdd

We can rearrange the second equation to state dd𝑥=17𝑢. Substituting this change of variable into our integral, we get (32𝑢48𝑢𝑢)17𝑢=327𝑢487𝑢𝑢𝑢=327𝑢𝑢487𝑢𝑢𝑢,cossectandcossectandcosdsectand

We apply the formulas above to obtain 327𝑢𝑢487𝑢𝑢𝑢=327(𝑢+)487(𝑢+)cosdsectandsinCsecC for some arbitrary constants C and C. Since, after distributing through the parenthesis, we will end up with a combination of C and C, we can replace this expression by another arbitrary constant C to write the solution as 327𝑢487𝑢+.sinsecC

Substituting 𝑢=7𝑥 back into the expression, we obtain the indefinite integral 87𝑥47𝑥+67𝑥𝑥=3277𝑥4877𝑥+.seccostandsinsecC

Let us consider another example involving the product of sec𝑥 and tan𝑥.

Example 3: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions

Determine 7𝑥(𝑥5𝑥)𝑥sectansecd.

Answer

Since there is a factored expression within the integrand, we should start by expanding through the parenthesis in the integrand. We obtain 7𝑥(𝑥5𝑥)=7𝑥𝑥35𝑥.sectansecsectansec

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: 𝑥𝑥𝑥=𝑥+,𝑥𝑥=𝑥+.sectandsecCsecdtanC

Applying these formulae to our integral, we obtain 7𝑥𝑥35𝑥𝑥=7𝑥𝑥𝑥35𝑥𝑥=7(𝑥+)35(𝑥+)sectansecdsectandsecdsecCtanC for some arbitrary constants C and C. Since, after distributing through the parenthesis, we will end up with a combination of C and C, we can replace this expression by another arbitrary constant C to write the solution as 7𝑥(𝑥5𝑥)𝑥=7𝑥35𝑥+.sectansecdsectanC

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 csc𝑥 and cot𝑥. Recall that ddcscdcsccot𝑥𝑥𝑥=𝑥𝑥.

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

𝑥𝑥𝑥=𝑥+csccotdcscC

To help with remembering this formula, we note the resemblance between this formula and the first formula we stated, which is 𝑥𝑥𝑥=𝑥+.sectandsecC

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, cossincottancscsec𝑥=𝜋2𝑥,𝑥=𝜋2𝑥,𝑥=𝜋2𝑥.

Therefore, we can write 𝑥𝑥𝑥=𝜋2𝑥𝜋2𝑥𝑥.csccotdsectand

We can use the 𝑢-substitution method by defining 𝑢=𝜋2𝑥,𝑢=𝑥.whichleadstodd

We can write the last equation as dd𝑥=𝑢. Substituting this change of variable into our indefinite integral, we get 𝜋2𝑥𝜋2𝑥𝑥=𝑢𝑢𝑢.sectandsectand

Applying the formula for the antiderivative of sectan𝑢𝑢, we obtain 𝑢𝑢𝑢=𝑢+.sectandsecC

Replacing 𝑢=𝜋2𝑥 back into the resulting expression, we obtain =𝜋2𝑥+=𝑥+.secCcscC

We note that, when integrating complementary trigonometric functions, we expect a negative sign resulting from the 𝑢 substituting 𝑢=𝜋2𝑥, which leads to dd𝑢=𝑥. 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.

Example 4: Finding the Integration of a Function Involving Trigonometric Functions

Determine 23𝑥3𝑥𝑥csccotd.

Answer

The given integrand is the product of the cosecant and cotangent functions. Hence, we recall the following integral: 𝑥𝑥𝑥=𝑥+.csccotdcscC

Before applying this formula to our integral, let us use the 𝑢-substitution method by defining 𝑢=3𝑥,𝑢=3𝑥.whichimpliesdd

We can also write the last equation as dd𝑥=13𝑢. Substituting this change of variables to our integral, we have 23𝑥3𝑥𝑥=23𝑥3𝑥13𝑥=23𝑢𝑢𝑢.csccotdcsccotdcsccotd

We can now apply our formula we recalled earlier to write this indefinite integral as 23(𝑢+)=23𝑢+23.cscCcscC

Since C is an arbitrary constant, we can simply denote the constant 23C as C in our answer. Substituting 𝑢=3𝑥 back into the resulting expression, we obtain 23𝑥3𝑥𝑥=233𝑥+.csccotdcscC

For our final formula, we state the indefinite integral of csc𝑥. To obtain this integral, we first recall the integral of sec𝑥: 𝑥𝑥=𝑥+.secdtanC

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

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

𝑥𝑥=𝑥+cscdcotC

When integrating tan𝑥 or cot𝑥, we can use the following trigonometric identity to express the function in terms of sec𝑥 or csc𝑥 respectively: tanseccotcsc𝑥+1=𝑥,𝑥+1=𝑥.

Both of these identities can be derived from the Pythagorean identity sincos𝑥+𝑥=1 when both sides of the equation are divided by either sin𝑥 or cos𝑥.

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

Example 5: Integrating a Reciprocal Trigonometric Function Whose Argument Has the Form 𝑎𝑥 + 𝑏

Determine 5(4𝑥+7)+1𝑥cotd.

Answer

Since 5 is a constant, we can begin by bringing it outside the integral: 5(4𝑥+7)+1𝑥.cotd

We can use the 𝑢-substitution method by defining 𝑢=4𝑥+7,𝑢=4𝑥.whichleadstodd

We can also write the last equation as dd𝑥=14𝑢. Substituting this change of variables to our integral, we obtain 5𝑢+114𝑥=54𝑢+1𝑥.cotdcotd

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

To obtain the relation between cot𝑥 and csc𝑥, we divide both sides of the Pythagorean identity sincos𝑥+𝑥=1 by sin𝑥: 1+𝑥𝑥=1𝑥.cossinsin

This is the same as cotcsc𝑥+1=𝑥.

We note that our integrand is precisely cot𝑢+1. Hence, applying this identity to our integral, we get 54𝑢𝑢.cscd

We can now apply the formula for the antiderivative of csc𝑢 to evaluate this indefinite integral: 54(𝑢+)=54𝑢54.cotCcotC

We can replace 54C by C since C is an arbitrary constant. We can also replace 𝑢 by 4𝑥+7. Hence, 5(4𝑥+7)+1=54(4𝑥+7)+.cotcotC

Let us recap a few important concepts from this explainer.

Key Points

  • The indefinite integral of the product of sec𝑥 and tan𝑥 is 𝑥𝑥𝑥=𝑥+.sectandsecC
  • 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 𝑥𝑥𝑥=𝑥+,𝑥𝑥=𝑥+.csccotdcscCcscdcotC
  • Integrals of tan𝑥 or cot𝑥 are not readily available. To compute an integral involving either of these functions, we can use the following trigonometric identities to relate it to sec𝑥 or csc𝑥 respectively: tanseccotcsc𝑥+1=𝑥,𝑥+1=𝑥.

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