# Trigonometric Integrals Explained [What are Trigonometric Integrals?]

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To understand the concept of trigonometric integrals, we must know what trigonometry and integrals are.

## Origin of Trigonometry

HIPPARCHUS of NICAEA is recognized as the father of trigonometry. He discovered the first trigonometric table.

Trigonometry is derived from a Greek word ‘trigonon’ which means ‘triangle’ and ‘metron’ (which means to measure). In this branch of mathematics, we study relationships between side lengths and angles of a triangle.

In my opinion, “when an individual starts studying trigonometry, he/she finds it difficult to understand as it consists of complex identities. But if these identities are well known to the individual, then all the problems will be solved easily and will start enjoying it.”

## Use of Integrals

Integration is one of the most beautiful parts of calculus, in my opinion. Because of its robust problem-solving methodology that uncovered a uniform approach to solving an enormous variety of complicated problems.

In my perspective, solving trigonometric integrals is the best exercise for our brain. It increases the thinking ability. It is not hard once you have enough practice.

To solve problems of trigonometric integrals, one must go through a variety of identities and formulas are given ahead.

## Division of Trigonometric Integrals

The Division of Trigonometric integrals are as follows:

1. Sine Integral
2. Cosine Integral
3. Hyperbolic Sine Integral Hyperbolic Cosine Integral
4. Auxiliary Functions
5. Nielsen’s Spiral
6. Exponential Integral

## Basic Useful Formulas

1. sin^2(x) + cos^2(x) = 1
2. tan^2(x) + 1 = sec^2(x)s
3. sin(2x) = 2 sin(x) cos(x)
4. cos(2x) = cos^2(x) − sin^2(x).

## Trigonometric Integrals

Here, we will look at various methods to integrate functions which are products and quotients of trigonometric functions. The starting point is the integrals of sine and cosine.

∫ sin x dx  =  – cos x  …………(1)

∫ cos x dx  =  sin x  …………..(2)

Next, come the trigonometric integrals of tangent and co-tangent

### Example 1:

Find ∫ tan x dx

Solution:

Since tan x = one can substitute u = cos x with du = – sin x dx to get

∫ tan x dx  =  ∫  dx  =  – ∫  du  =  – ln | u |  =  – ln | cos x |  =  ln

So

∫ tan x dx  =  ln | sec x | …………..(3)

If we are working in an interval where cos x is positive then we can just write tan x dx  =  ln sec x.

Similarly

∫ cot x dx  =  ln | sin x | …………..(4)

Finally, we come to the integrals of secant and co-secant. The easiest way to do these is to multiply top and bottom by sec x + tan x or cosec x + cot x.

### Example 2

Find ∫ sec x dx

Solution:

Multiply the bottom and top by sec x + tan x and then substitute u = tan x + sec x with du = (sec^2(x) + tan x sec x) dx to get

∫ sec x dx  =   dx  =   dx  =   du  =  – ln | u |

=  ln | tan x + sec x

So

∫ sec x dx  =  ln | tan x + sec x…………..(5)

If we are working in an interval where tan x + sec xtan x + sec x is positive then we can just write

∫ sec x dx  =  ln (tan x + sec x).

Similarly

∫ cosec x dx  =  – ln | cot x + cosec x | …………..(6)

Now we look at some techniques for integrating products of powers of sines and cosines, i.e., ∫ sin^n(x) cos^m(x) dx

## Case 1: One Power is Odd and Positive

The simplest example is when one of the powers is an odd positive integer. Suppose for definiteness that m is odd. In that case, we split off a cos x and use it with the dx in the substitution u = sin x. We express the remaining cosine terms in terms of sine using sin^2(x) + cos^2(x) = 1 for the substitution.

Let’s look at an example.

### Example 3

Find ∫ dx.

Solution:

∫  dx  =  ∫  cos x dx  =  ∫  cos x dx  =  ∫  cos x dx

Now let u = sin x with du = cos x dx. We get

∫  dx  =  ∫  du

Now expand the top and split up the fraction.

∫  du  =  ∫  du  =  ∫ ( – + )du  =  ∫ (u^-6 – 2(u^-4) + u^-2 )du

=  – cosec5x + cosec3x – cosecx

This is how you solve Trigonometric Integrals.

Problem:  Find ∫ sin^3(x) cos^5(x) dx

## Case 2: Both Powers are Even

In this case use the following identities to reduce the powers, substitute u = 2x and reassess the situation.

sin 2x = 2 sinx cosx                 sin2x =              cos2x =

### Example 4

∫ sin^2(x) cos^4(x) dx.

Solution:

∫ sin^2(x) cos^4(x) dx =  ∫ (sin x cos x)^2 (cos^2(x)) dx =  ∫ ^2 dx

Now let u = 2x with du = dx

The integral becomes

∫ sin^2u (1 + cos u) du  =   ∫ sin^2u du +  ∫ sin^2u cos u du  …………..(7)

In the first integral use the second identity in (7). In the second, let v = sin u

The integral becomes

∫  du +  ∫ v^2 dv

Problem: ∫ sin^2 x dx and ∫ sin^4 x dx

We can do similar things, for integrating products of powers of tangents and secants i.e.,

∫ tan^nx sec^mx dx

This is how you solve Trigonometric Integrals

## Case 3: ‘tan’ is to an Odd Positive Power

In that case, we split off a tan x and use it with sec x dx in the substitution u = sec x. We express the remaining tangent terms in terms of secant using sec^2x – tan^2x = 1 for the substitution.

Let’s look at an example.

### Example 3

Find ∫ tan^3 x sec^3 x dx

Solution:

∫ tan^3 x sec^3 xdx  =  ∫ tan^2 x sec^2 x tan x sec xdx

=  ∫ (sec^2 x – 1) sec^2 x tan x sec x dx

Let u = sec x with du = tan x sec x dx

The integral becomes

∫ (u^2 – 1) u^2 du  =  ∫ (u^4 – u^2) du

This is how you can solve different types of Trigonometric Integrals.

Hello folks. I have started this website for all of you who have troubles grasping those tough mathematical concepts. I personally have faced a lot of issues and that is the sole motivation for starting this. Hope it helps you guys. Cheers!