Integration of Trigonometric Functions (original) (raw)

Last Updated : 2 Dec, 2025

Integration is the process of summing up small values of a function in the region of limits. It is just the opposite of differentiation. Integration is also known as an anti-derivative.

Below is an example of the Integration of a given function.

**e.g., Consider a function, f(y) = y2.

This function can be integrated as:

∫y2dy = \frac{y^{2+1}}{2+1}~+~C

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:

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

Generally, the problems of indefinite integrals based on trigonometric functions are solved by the substitution method.

**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 ∫ 3x 2 sin (x 3 ) 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 = ∫ 3x 2 sin (x 3 ) dx = -cos x 3 + 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:

Sample Problems on Integration of Trigonometric Functions

**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, ∫ cos 3 x dx = sin x - sin 3 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 = sin 3 x / 3 - sin 5 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, ∫ sin 4 (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.