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Integration of Trigonometric Functions

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Differentiation, in mathematics, the process of finding the derivative, or rate of change, of a function. In contrast, to the abstract nature of the theory behind it, the practical technique of differentiation can be carried out by purely algebraic manipulations, using three basic derivatives, four rules of operation, and a knowledge of how to manipulate functions.

e.g.: Consider a function, x = y2

This function can be differentiated as: 

d(y2) / dy = 2y 

However, an indefinite integral is a function that takes the anti-derivative of another function. It is represented as an integral symbol (∫), a function, and a derivative of the function at the end. The indefinite integral is an easier way to symbolize an anti-derivative.

Let’s learn what is integration mathematically, the integration of a function f(x) is given by F(x) and it is represented by:

∫f(x)dx = F(x) + C

Here R.H.S. of the equation means integral of f(x) with respect to x, F(x) is called anti-derivative or primitive, f(x) is called the integrand, dx is called the integrating agent, C is called constant of integration or arbitrary constant and x is the variable of integration.

Some Important Integrals of Trigonometric Functions

Following is the list of some important formulae of indefinite integrals on basic trigonometric functions to be remembered as follows:

  • ∫ sin x dx = -cos x + C
  • ∫ cos x dx = sin x + C
  • ∫ sec2 x dx = tan x + C
  • ∫ cosec2 x dx = -cot x + C
  • ∫ sec x tan x dx = sec x + C
  • ∫ cosec x cot x dx = -cosec x + C
  • ∫ tan x dx = ln | sec x | + C
  • ∫ cot x dx = ln | sin x | + C
  • ∫ sec x dx = ln | sec x + tan x | + C
  • ∫ cosec x dx = ln | cosec x – cot x | + C

where dx is the derivative of x, C is the constant of integration and ln represents the logarithm of the function inside modulus (| |).

Generally, the problems of indefinite integrals based on trigonometric functions are solved by the substitution method. So let’s discuss more about the integration by substitution method as follows:

Integration by Substitution

In this method of integration by substitution, any given integral is transformed into a simple form of integral by substituting the independent variable with others. Let’s consider an example for better understanding.

Example: Simplify ∫ 3x2 sin (x3) dx.

Answer:

Let I = ∫ 3x2 sin (x3) dx.

In order to evaluate the given integral lets substitute any variable by a new variable as:

Let x3 be t for the given integral.

Then, dt = 3x2 dx

Therefore, 

I = ∫ 3x2 sin (x3) dx = ∫ sin (x3) (3x2 dx)

Now, substitute t for x3 and dt for 3x2 dx in the above integral.

I = ∫ sin (t) (dt)

As ∫ sin x dx = -cos x + C, thus

I = -cos t + C

Again, substitute back x3 for t in the expression as:

I = ∫ 3x2 sin (x3) dx = -cos x3 + C

Which is the required integral.

Hence, the General Form of integration by substitution is:

∫ f(g(x)).g'(x).dx = f(t).dx

Where t = g(x)

Usually, the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. By doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.

In calculus, the integration by substitution method is also known as the “Reverse Chain Rule” or “U-Substitution Method”. We can use this method to find an integral value when it is set up in the special form. It means that the given integral is of the form:

Now, let’s solve some basic problems based on the concepts discussed above as follows:

Sample Problems

Problem 1: Determine the integral of the following function: f(x) = cos3 x.

Solution:

Let us consider the integral of the given function as,

I = ∫ cos3 x dx

It can be rewritten as:

I = ∫ (cos x) (cos2x) dx

Using trigonometry identity; cos2x = 1 – sin2x, we get

I = ∫ (cos x) (1 – sin2x) dx

⇒ I = ∫ cos x – cos x sin2x dx 

⇒ I = ∫ cosx dx – ∫ cosx sin2x dx

As ∫ cos x dx = sin x + C,

Thus, I = sin x – ∫ sin2x cos x dx                         . . . (1)

Let, sin x = t 

⇒ cos x dx = dt.

Substitute t for sin x and dt for cos x dx in second term of the above integral.

I = sin x – ∫ t2 dt

⇒ I = sin x – t3/3 + C

Again, substitute back sin x for t in the expression.

Hence, ∫ cos3x dx = sin x – sin3 x / 3 + C.

Problem 2: If f(x) = sin2 (x) cos3 (x) then determine ∫ sin2(x) cos3(x) dx.

Solution:

Let us consider the integral of the given function as,

I = ∫ sin2(x) cos3(x) dx

Using trigonometry identity; cos2 x = 1 – sin2 x, we get

I = ∫ sin2 x (1 – sin2 x) cos x dx

Let sin x = t then,

⇒ dt = cos x dx

Substitute these in the above integral as,

I = ∫ t2 (1 – t2) dt

⇒ I = ∫ t2 – t4 dt

⇒ I = t3 / 3 – t5 / 5 + C  

Substitute back the value of t in the above integral as,

Hence, I = sin3 x / 3 – sin5 x / 5 + C.

Problem 3: Let f(x) = sin4(x) then find ∫ f(x)dx. i.e. ∫ sin4(x) dx.

Solution:

Let us consider the integral of the given function as,

I = ∫ sin4(x) dx

⇒ I = ∫ (sin2(x))2 dx

Use\ing trigonometry identity; sin2(x) = (1 – cos (2x)) / 2, we get

I = ∫ {(1 – cos (2x)) / 2}2 dx

⇒ I = (1/4) × ∫ (1+cos2(2x)- 2 cos2x) dx

⇒ I = (1/4) × ∫ 1 dx + ∫ cos2(2x) dx – 2 ∫ cos2x dx

⇒ I = (1/4) × [ x + ∫ (1 + cos 4x) / 2 dx – 2 ∫ cos2x dx ] 

⇒ I = (1/4) × [ 3x / 2 + sin 4x / 8 – sin 2x ] + C

⇒ I = 3x / 8 + sin 4x / 32 – sin 2x / 4 + C

Hence, ∫ sin4(x) dx = 3x / 8 + sin 4x / 32 – sin 2x / 4 + C

Problem 4: Find the integration of \bold{\int\frac{e^{tan^{-1}x}}{1+x^2} dx}    .

Solution:

Let us consider the integral of the given function as,

I =\int \frac{e^{tan^{-1}x}}{1+x^2} dx

Let t = tan-1 x                         . . . (1)

Now, differentiate both side with respect to x:

dt = 1 / (1+x2) dx

Therefore, the given integral becomes:

I = ∫ et dt

⇒ I = et + C                               . . . (2)

Substitute the value of (1) in (2) as:

⇒  I = e^{tan^{-1}x} + C

Which is the required integration for the given function.

Problem 5: Find the integral of the function f (x) defined as,

f(x) = 2x cos (x2 – 5) dx

Solution:

Let us consider the integral of the given function as,

I = ∫ 2x cos (x2 – 5) dx

Let (x2 – 5) = t                         . . . (1)

Now differentiate both side with respect to x as,

2x dx = dt 

Substituting these values in the above integral,

I = ∫ cos (t) dt

⇒ I = sin t + C                         . . . (2)

Substitute the value equation (1) in equation (2) as,

⇒ I = sin (x2 – 5) + C

This is the required integration for the given function. 

Problem 6: Determine the value of the given indefinite integral, I = ∫ cot (3x +5) dx.

Solution:

The given integral can be written as,

I = ∫ cot (3x +5) dx 

⇒ I = ∫ cos (3x +5) / sin (3x +5) dx 

Let, t = sin(3x + 5)

⇒ dt = 3 cos (3x+5) dx

⇒ cos (3x+5) dx = dt / 3 

Thus,

I = ∫ dt / 3 sin t  

⇒ I = (1 / 3) ln | t | + C

Replace t by sin (3x+5) in the above expression.

I = (1 / 3) ln | sin (3x+5) | + C

This is the required integration for the given function.

FAQs on Integration of Trigonometric Functions

Q1: What is the Integration of a Trigonometric Function?

Answer:

The integration of trigonometric functions as the name suggests is the process of calculating the integration or antiderivative of trigonometric functions. This is the reverse process of differentiation of trigonometric functions.

Q2: What are Basic Trigonometric Functions?

Answer:

The basic trigonometric functions are: 

  • sine (sin), 
  • cosine (cos), 
  • tangent (tan), 
  • cotangent (cot), 
  • secant (sec), and 
  • cosecant (csc).

Q3: How do you Integrate Sine (sin) and Cosine (cos) functions?

Answer:

To integrate the sine and cosine functions, we can use the following formulas:

  • ∫ sin(x) dx = -cos(x) + C
  • ∫ cos(x) dx = sin(x) + C

Where C is the constant of integration.

Q4: What is the Integration of the Tangent (tan) Trigonometric Function?

Answer:

The integral of the tangent function is given as follows:

∫ tan(x) dx = -ln|cos(x)| + C

Where,

  • ln represents the natural logarithm, and 
  • C is the constant of integration.

Q5: How to Find the Integral of the Secant (Sec) Trigonometric Function?

Answer:

The integral of the secant function is given as:

∫ sec(x) dx = ln|sec(x) + tan(x)| + C

Where,

  • ln represents the natural logarithm, and 
  • C is the constant of integration.

Q6: What is the Integration of the Cotangent (cot) Trigonometric Function?

Answer:

The integral of the cotangent function can be calcualted using the following formula:

∫ cot(x) dx = ln|sin(x)| + C

Where,

  • ln represents the natural logarithm, and 
  • C is the constant of integration.

Q7: How to Find the Integral of the Cosecant (cosec) Function?

Answer:

The integral of the cosecant function is given as:

∫ cosec(x) dx = ln| cosec x – cot x | + C

Where,

  • ln represents the natural logarithm, and 
  • C is the constant of integration.


Last Updated : 24 Jul, 2023
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