Integration of Trigonometric Functions

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.

List of some important Indefinite Integrals of Trigonometric Functions

Following is the list of some important formulae of indefinite integrals on basic trigonometric functions to be remembered are 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 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 by others.

Consider an integral, ∫ 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, 

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

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

∫ 3x² sin (x³) dx = ∫ sin (t) (dt)

                          = -cos t + C                                                                   (Since, ∫ sin x dx = -cos x + C)

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

∫ 3x² sin (x³) dx = -cos x³ + C

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. 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, lets 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

Use trigonometry identity: cos²x = 1 – sin²x as,

I = ∫ (cos x) (1 – sin²x) dx

  = ∫ cos x – cos x sin²x dx 

  = ∫ cosx dx – ∫ cosx sin²x dx

  = sin x – ∫ sin²x cos x dx.                                                         (Since, ∫ cos x dx = sin x + C)       ……(1)

Let, sin x = t then, 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

  = 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

Use trigonometry identity: cos² x = 1 – sin² x as,

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

  = ∫ t² – t⁴ dt

  = 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

or

I = ∫ (sin²(x))² dx

Use trigonometry identity: sin²(x) = (1 – cos (2x)) / 2 as,

I = ∫ {(1 – cos (2x)) / 2}² dx

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

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

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

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

  = 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:

Solution:

Let us consider the integral of the given function as,

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 = ∫ e^t dt

  = e^t + C                                                                                                  …….(2)

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

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

  = 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 

  = ∫ cos (3x +5) / sin (3x +5) dx 

Let, t = sin(3x + 5) so,

dt = 3 cos (3x+5) dx

Therefore, 

cos (3x+5) dx = dt / 3 

And

I = ∫ dt / 3 sin t  

  = (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.