Lesson Explainer: Indefinite Integrals: Trigonometric Functions Mathematics • Higher Education

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

To find the indefinite integral of various trigonometric functions, we can start by recalling the first part of the fundamental theorem of calculus.

Theorem: The Fundamental Theorem of Calculus

If 𝑓 is a continuous real-valued function on an interval [π‘Ž,𝑏] and we let 𝐹 be defined on the interval [π‘Ž,𝑏] as 𝐹(π‘₯)=𝑓(𝑑)𝑑,ο—οŒΊd then 𝐹 is uniformly continuous on [π‘Ž,𝑏] and differentiable on (π‘Ž,𝑏), and 𝐹′(π‘₯)=𝑓(π‘₯), for all π‘₯∈(π‘Ž,𝑏).

The fundamental theorem of calculus tells us that differentiation and indefinite integration are the reverse processes of one another. This means we can find the indefinite integral of some trigonometric functions by considering the derivative results we already know.

For example, we know that, for any real constant 𝑛 and variable π‘₯ measured in radians, ddsincosπ‘₯((𝑛π‘₯))=𝑛(𝑛π‘₯).

This tells us that sin(𝑛π‘₯) is an antiderivative of 𝑛(𝑛π‘₯)cos, for any value of 𝑛. We can use this to find the most general antiderivative of 𝑛(𝑛π‘₯)cos; since this will be an antiderivative if we add any constant, we need to add a constant of integration we will call 𝑐: 𝑛(𝑛π‘₯)π‘₯=(𝑛π‘₯)+𝑐.cosdsin

This is a useful result; however, we can rearrange this equation to find an integral rule for cos(𝑛π‘₯) by using our properties of indefinite integrals: 𝑛(𝑛π‘₯)π‘₯=𝑛(𝑛π‘₯)π‘₯.cosdcosd

This means 𝑛(𝑛π‘₯)π‘₯=(𝑛π‘₯)+𝑐.cosdsin

We can then divide through by 𝑛, provided it is nonzero: ο„Έ(𝑛π‘₯)π‘₯=(𝑛π‘₯)𝑛+𝑐𝑛.cosdsin

Finally, 𝑐 is a constant, which means 𝑐𝑛 is also just a constant. This means we can introduce a new constant 𝐢=𝑐𝑛 to simplify this expression.

We have shown, provided 𝑛≠0, that ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢.cosdsin

We can follow exactly the same process to find an indefinite integral rule for sin(𝑛π‘₯).

First, we know that, for any real constant 𝑛 and variable π‘₯ measured in radians, ddcossinπ‘₯((𝑛π‘₯))=βˆ’π‘›(𝑛π‘₯).

This means cos(𝑛π‘₯) is an antiderivative of βˆ’π‘›(𝑛π‘₯)sin, and we can use this to find the most general antiderivative: ο„Έβˆ’π‘›(𝑛π‘₯)π‘₯=(𝑛π‘₯)+𝑐.sindcos

We can take the constant factor of βˆ’π‘› outside of our integral: βˆ’π‘›ο„Έ(𝑛π‘₯)π‘₯=(𝑛π‘₯)+𝑐.sindcos

We can then divide through by βˆ’π‘›, provided 𝑛≠0: ο„Έ(𝑛π‘₯)π‘₯=βˆ’1𝑛(𝑛π‘₯)+π‘βˆ’π‘›.sindcos

Finally, we set 𝐢=π‘βˆ’π‘›: ο„Έ(𝑛π‘₯)π‘₯=βˆ’1𝑛(𝑛π‘₯)+𝐢.sindcos

We will show one more example of this process on the tangent function.

We recall that, for any real constant 𝑛 and variable π‘₯ measured in radians, ddtansecπ‘₯((𝑛π‘₯))=𝑛(𝑛π‘₯).

This tells us that tan(𝑛π‘₯) is an antiderivative of 𝑛(𝑛π‘₯)sec. Hence, we can use this to find the most general antiderivative: 𝑛(𝑛π‘₯)π‘₯=(𝑛π‘₯)+𝑐.secdtan

Taking the constant factor 𝑛 outside of the indefinite integral gives us 𝑛(𝑛π‘₯)π‘₯=𝑛(𝑛π‘₯)π‘₯.secdsecd

Then, provided 𝑛≠0, we can divide through by 𝑛: ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢,secdtan where 𝐢=𝑐𝑛.

Let’s summarize the indefinite integral results we have just shown.

Definition: Indefinite Integrals Resulting in Trigonometric Functions

For any real constant 𝑛≠0 and variable π‘₯ measured in radians,

  • ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢cosdsin,
  • ο„Έ(𝑛π‘₯)π‘₯=βˆ’1𝑛(𝑛π‘₯)+𝐢sindcos,
  • ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢secdtan.

One way of visualizing this relationship is to use the following diagram.

Differentiating with respect to π‘₯ takes us clockwise through this cycle and integrating with respect to π‘₯ will take us counterclockwise, where we will need to add a constant of integration.

Let’s look at a few examples to practice and help strengthen our understanding. In our first example, we will demonstrate how to evaluate the indefinite integral of a sum of two trigonometric functions.

Example 1: Integrating Trigonometric Functions

Determine ο„Έ(9π‘₯+4π‘₯)π‘₯.sincosd

Answer

We want to evaluate the indefinite integral of a trigonometric function. We can start by simplifying this integral using our properties of indefinite integration: ο„Έ(9π‘₯+4π‘₯)π‘₯=ο„Έ9π‘₯π‘₯+ο„Έ4π‘₯π‘₯=9ο„Έπ‘₯π‘₯+4ο„Έπ‘₯π‘₯.sincosdsindcosdsindcosd

We can then evaluate these indefinite integrals by recalling ο„Έπ‘₯π‘₯=βˆ’π‘₯+𝐴,ο„Έπ‘₯π‘₯=π‘₯+𝐡.sindcoscosdsin

Combining all of the constants of integration at the end of our expression, this gives us 9ο„Έπ‘₯π‘₯+4ο„Έπ‘₯π‘₯=9(βˆ’π‘₯+𝐴)+4(π‘₯+𝐡)=βˆ’9π‘₯+4π‘₯+9𝐴+4𝐡=βˆ’9π‘₯+4π‘₯+𝐢.sindcosdcossincossincossin

Hence, ο„Έ(9π‘₯+4π‘₯)π‘₯=4π‘₯βˆ’9π‘₯+𝐢.sincosdsincos

In our next example, we will see how to apply this process to the indefinite integral of a trigonometric function involving multiple angles.

Example 2: Integrating Trigonometric Functions with Multiple Angles

Determine ο„Έ36π‘₯π‘₯.cosd

Answer

To evaluate this indefinite integral, we start by recalling the following indefinite integral result for trigonometric functions. For any real constant 𝑛≠0, ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢.cosdsin

To apply this, we set 𝑛=6 and we take the constant factor 3 outside of our integral: ο„Έ36π‘₯π‘₯=3ο„Έ6π‘₯π‘₯=3ο€½(6π‘₯)6+𝑐=3(6π‘₯)6+3𝑐=(6π‘₯)2+3𝑐.cosdcosdsinsinsin

Finally, we will set 𝐢=3𝑐 and rewrite the factor of 12 in front of our function, giving us sinsin(6π‘₯)2+3𝑐=12(6π‘₯)+𝐢.

Hence, ο„Έ36π‘₯π‘₯=126π‘₯+𝐢.cosdsin

We can also use these rules for indefinite integrals to evaluate integrals involving the antiderivative of the square of the secant function.

Example 3: Integrating Reciprocal Trigonometric Functions

Determine ο„Έβˆ’6π‘₯π‘₯.secd

Answer

To evaluate this indefinite integral, we first recall the following indefinite integral result for trigonometric functions. For any real constant 𝑛≠0, ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢.secdtan

To apply this result to the indefinite integral in the question, we will first take the constant factor βˆ’1 outside of our integral: ο„Έβˆ’6π‘₯π‘₯=βˆ’ο„Έ6π‘₯π‘₯.secdsecd

We then apply our indefinite integral rule with 𝑛=6: βˆ’ο„Έ6π‘₯π‘₯=βˆ’ο€½(6π‘₯)6+𝑐.secdtan

Finally, we will introduce a new constant 𝐢=βˆ’π‘ and write the coefficient of βˆ’16 at the front of the function: βˆ’ο€½(6π‘₯)6+𝑐=βˆ’16(6π‘₯)+𝐢.tantan

Therefore, we have shown ο„Έβˆ’6π‘₯π‘₯=βˆ’166π‘₯+𝐢.secdtan

We can also apply these results when the coefficient of our variable is not an integer. We will demonstrate this in our next example.

Example 4: Integrating Trigonometric Functions

Determine ο„Έ7ο€Ό2π‘₯3π‘₯.cosd

Answer

To help us evaluate this indefinite integral, it is easier to rewrite the argument as 23π‘₯: ο„Έ7ο€Ό2π‘₯3π‘₯=ο„Έ7ο€Ό23π‘₯π‘₯.cosdcosd

We can then evaluate this integral by recalling the following indefinite integral rules for trigonometric functions; for any real constant 𝑛≠0, ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢.cosdsin

We can use this to evaluate our indefinite integral by taking the constant factor 7 outside the integral and setting 𝑛=23. This gives us 7ο„Έο€Ό23π‘₯π‘₯=7π‘₯+𝑐=7ο€Ό32ο€Ό23π‘₯+π‘οˆ.cosdsinsin

Finally, we can multiply through by 7 and introduce a new constant 𝐢=7𝑐: 7ο€Ό32ο€Ό23π‘₯+π‘οˆ=212ο€Ό2π‘₯3+𝐢.sinsin

Therefore, ο„Έ7ο€Ό2π‘₯3π‘₯=212ο€Ό2π‘₯3+𝐢.cosdsin

In our next example, we will see how to use these rules of indefinite integration together with the power rule of integration.

Example 5: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions

Determine ο„Έο€Ήβˆ’67π‘₯+26π‘₯βˆ’1π‘₯.sinsecd

Answer

We are asked to evaluate the indefinite integral of the sum or difference of three terms. Let’s start by rewriting this as three separate integrals: ο„Έο€Ήβˆ’67π‘₯+26π‘₯βˆ’1π‘₯=ο„Έβˆ’67π‘₯π‘₯+ο„Έ26π‘₯π‘₯+ο„Έβˆ’1π‘₯=βˆ’6ο„Έ7π‘₯π‘₯+2ο„Έ6π‘₯π‘₯βˆ’ο„Έ1π‘₯.sinsecdsindsecddsindsecdd

We can evaluate each of these indefinite integrals separately by recalling the following three indefinite integral rules.

For any nonzero real constant π‘Ž, we have ο„Έ(π‘Žπ‘₯)π‘₯=βˆ’1π‘Ž(π‘Žπ‘₯)+𝑐,ο„Έ(π‘Žπ‘₯)π‘₯=1π‘Ž(π‘Žπ‘₯)+𝑐.sindcossecdtan

For any real constant π‘›β‰ βˆ’1, the power rule for integration tells us ο„Έπ‘₯π‘₯=π‘₯𝑛+1+𝑐.d

With 𝑛=0, this tells us ο„Έ1π‘₯=π‘₯+𝑐.d

We can use these three rules to integrate each term separately: βˆ’6ο„Έ7π‘₯π‘₯+2ο„Έ6π‘₯π‘₯βˆ’ο„Έ1π‘₯=βˆ’6ο€Όβˆ’7π‘₯7+𝑐+2ο€Ό6π‘₯6+π‘βˆ’π‘₯+𝑐.sindsecddcostan

Finally, we can combine all of the constants of integration into a new constant, 𝐢: βˆ’6ο€Όβˆ’7π‘₯7+𝑐+2ο€Ό6π‘₯6+π‘βˆ’π‘₯+𝑐=βˆ’π‘₯+677π‘₯+136π‘₯+𝐢.costancostan

Hence, ο„Έο€Ήβˆ’67π‘₯+26π‘₯βˆ’1π‘₯=βˆ’π‘₯+677π‘₯+136π‘₯+𝐢.sinsecdcostan

In our final example, we will combine one of the trigonometric identities with our indefinite integral rules.

Example 6: Integrating Reciprocal Trigonometric Functions

Determine ο„Έβˆ’549π‘₯π‘₯.cosd

Answer

In this question, we are asked to find the indefinite integral of a trigonometric function. However, in its current form, we do not know how to integrate this function directly. Instead, we can start by rewriting our integrand using the following reciprocal trigonometric identity: 1πœƒβ‰‘πœƒ.cossec

Using this identity and our rules for definite integrals, we get ο„Έβˆ’549π‘₯π‘₯=ο„Έβˆ’54ο€Ό19π‘₯π‘₯=ο„Έβˆ’54ο€Ό19π‘₯π‘₯=βˆ’54ο„Έ9π‘₯π‘₯.cosdcosdcosdsecd

We can now evaluate this indefinite integral by recalling the following rule. For any real constant π‘Žβ‰ 0, ο„Έ(π‘Žπ‘₯)π‘₯=1π‘Ž(π‘Žπ‘₯)+𝐢.secdtan

By setting π‘Ž=9, we have βˆ’54ο„Έ9π‘₯π‘₯=βˆ’54ο€Ό9π‘₯9+π‘οˆ.secdtan

Then, we expand and set 𝐢=βˆ’5𝑐4: βˆ’54ο€Ό9π‘₯9+π‘οˆ=βˆ’5369π‘₯+𝐢.tantan

Hence, ο„Έβˆ’549π‘₯π‘₯=βˆ’5369π‘₯+𝐢.cosdtan

Let’s finish by recapping some of the key points of this explainer.

Key Points

  • By using the fundamental theorem of calculus and the rules for differentiating trigonometric functions, we are able to demonstrate the following rules for finding the indefinite integral of trigonometric functions.
  • For any real constant 𝑛, not equal to zero, and variable π‘₯ measured in radians,
    • ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢cosdsin,
    • ο„Έ(𝑛π‘₯)π‘₯=βˆ’1𝑛(𝑛π‘₯)+𝐢sindcos,
    • ο„Έ(𝑛π‘₯)π‘₯=1𝑛(𝑛π‘₯)+𝐢secdtan.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.