Introduction

Integration refers to the computation of an integral. In mathematics, integrals help us in determining many quantities such as displacement, area, and volume, etc. Generally, when we are speaking about integrals, then it means we are referring to definite integrals. For antiderivatives, we employ indefinite integrals.  Integration and differentiation are two major concepts in calculus. Differentiation refers to the instantaneous rate of change of the function at a given time. Integration is the reverse process of differentiation.

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What are Trigonometric Functions?

Trigonometric functions, also known as circular functions, refer to the functions of an angle of a triangle. It means that these trigonometric functions tell us the relationship between the angles and sides of a triangle. The basic trig functions are given below:

  • Sine
  • Cosine
  • Tangent
  • Cotangent
  • Secant
  • Cosecant

In this article, we will specifically discuss how to compute the integrals of trigonometric functions.

 

Trigonometric Integrals

Trigonometric integrals refer to the integrals of trigonometric functions. Some of the common integrals of trigonometric functions are given below:

1. \int cos x dx = sin x + C

2. \int sin x dx = -cos x + C

3. \int sec^2 x dx = tan x + C

4. \int cosec^2 x dx = - cotanx + C

5. \int (sec x tan x)dx = sec x + C

6. \int (cosec x cotan x)dx = - cosec x + C

 

Techniques for Finding the Integrals of the Trigonometric Functions

In this section of the article, we will discuss what techniques we can use to find the integrals of the trigonometric functions of the form \int cos ^j sin ^k x dx.

Strategy 1

  • If the exponent k is an odd number, then we can rewrite sin^k x as sin ^{k - 1} x sin x. After that, we can use the trig identity rule sin^2 x = 1 - cos^2 x to rewrite sin ^{k - 1} x in terms of cos x. Next, we will integrate the function by substituting cos x = u.  This substitution will give us the value of du = -sin x dx.

Strategy 2

  • If the exponent j is an odd number, then we can  rewrite cos^j x as cos ^{j - 1} x cos x. In the next step, we will use the identity rule cos^2 x = 1 - sin^2 x to rewrite cos ^{j - 1} x in terms of sin x. Next, we will  integrate the function by substituting sin x = u.  This substitution will provide us the value of du = cos x dx.

Strategy 3

  • If both the exponents j and k are odd numbers, then we can solve the question by using either the first or the second strategy.

Strategy 4

  • If both the exponents  j and k are even numbers, then we will use sin^2 x = \frac{1}{2} - \frac{1}{2} cos (2x), and cos^2x = (\frac{1}{2}) +( \frac{1}{2} cos (2x). After we have applied these formulas, we can integrate the function by applying the above strategies where necessary.

 

Now, we will solve some examples in which we will find the integrals of trigonometric functions.

 

Example 1

Calculate the integral of the following function:

\int (12 cosec^2 x + 5 sec^2 x) dx

Solution

\int (12 cosec^2 x + 5 sec^2 x) dx can be written as:

= \int 12 cosec^2 x  + \int 5 sec^2 x dx

According to one of the properties of integrals, we can shift the constants before the integral sign. Hence, we can rewrite the above function as:

= 12\int  cosec^2 x + 5 \int  sec^2 x dx

As we know that \int cosec^2 x dx = - cot x + C and \int sec^2 x dx = tan x + C. Hence, we can substitute these values in the above function to get the final answer:

=-12 cot x + 5 tan x + C

 

Example 2

Find the integral of the following function:

\int cos (13t + 9) dt

Solution

In this example, we will use substitution to find the integral. Suppose 13t + 9 = u.

If u = 13t + 9, then \frac{du}{dt} = 13. This means that dt = \frac{1}{13} du. We will substitute these values in the original function to get the following function:

= \int cos u \frac{1}{13} du

Shift the fraction before the integral sign:

= \frac{1}{13} \int cos u du

We know that \int cos x dx = tan x + C. Hence, we will substitute this value in the above function:

= \frac{1}{13}  sin u + C

Since u = 13t + 9, hence substitute this value with u in the above function again:

=  \frac{1}{13} sin (13t + 9) + C

 

 

Example 3

Find the integral of the following function:

\int sec^2 (6x - 3) dx

Solution

In this example, we will use substitution to find the integral. Suppose 6x - 3 = u.

If u = 6x - 3, then \frac{du}{dx} = 6. This means that dx = \frac{1}{6} du. We will substitute these values in the original function to get the following function:

= \int sec^2 u \frac{1}{6} du

Shift the fraction before the integral sign:

= \frac{1}{6} \int sec^2 u du

We know that \int sec^2 x dx = tan x + C. Hence, we will substitute this value in the above function:

= \frac{1}{6}  tan u + C

Since u = 6x - 3, hence substitute this value with u in the above function again:

=  \frac{1}{6} tan (6x - 3) + C

 

Example 4

Find the integral of the following function:

\int cos ^ 8 x  sin^3 x dx

Solution

You can see that the exponent of sin x is an odd number, hence we can rewrite the above function using strategy 1 like this:

= \int cos ^8 x sin^2x sin x dx

Substitute sin^2 x = 1 - cos^2 x in the above function like shown below:

= \int cos^8 (1 - cos^2x) sin x dx

Suppose u = cos x, then \frac {du}{dx} = sin x, and -du = sin x dx:

= \int u^8 (1 - u^2x) (-du)

= \int - (u^8 - u^{10})

= \int - u^8 + u^{10}

Integrate the above function like this:

= - \frac{1}{9} u^9 + \frac{1}{11} u^{11} + C

Substitute u = cos x again in the above function:

= - \frac{1}{9} cos ^9 x + \frac{1}{11} cos^{11} x + C

 

Example 5

Find the integral of the following function:

\int cos ^ 3 x  sin^5 x dx

Solution

You can see that the exponents of both the cos and sine functions are odd numbers, hence we can rewrite the above function either a strategy 1 or 2. In this example, we have chosen the strategy 2:

= \int cos ^2 x  cos x sin^5 xdx

Substitute cos^2 x = 1 - sin^2 x in the above function like shown below:

= \int (1 - sin^2 x) cos x sin^5 x dx

Suppose u = sin x, then \frac {du}{dx} = cos x, and du = cos x dx:

= \int (1 - u^2x) u^5(du)

= \int (u^5 - u^7) du

= \int u^5 du - \int u^7 du

Integrate the above function like this:

= \frac{1}{6} u^6 - \frac{1}{8} u^8 + C

Substitute u = cos x again in the above function:

= \frac{1}{6} sin ^6 x - \frac{1}{8} sin^8 x + C

 

 

Example 6

Find the integral of the following function:

\int sec^4 x  tan^9 x dx

Solution

Since sec^4 x = (sec^2 x)(sec^2 x), hence we can rewrite the above function in this way:

= \int sec ^2 x  sec^2 x tan^ 9 xdx

Substitute sec^2x = tan^2 x + 1 in the above function like shown below:

= \int (tan^2 x + 1) sec^2 tan^ 9 x dx

Suppose u = tan x, then \frac {du}{dx} = sec^2x, and du = sec^2 x dx:

= \int (1 + u^2x) u^9(du)

= \int (u^9 + u^{11})

= \int u^9 + \int u^{11}

Integrate the above function like this:

= \frac{1}{10} u^{10} + \frac{1}{12} u^{12} + C

Substitute u = tan x again in the above function:

= \frac{1}{10} tan ^ {10} x + \frac{1}{12} tan ^ {12} x + C

 

 

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Rafia Shabbir