Higher Level Mathematics Exploration Exploring the ... · HL Mathematics syllabus, making this is...

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Higher Level Mathematics Exploration

Exploring the techniques of integrating trigonometric functions, with strong

focus on applying Euler’s formula.

Candidate number: glt079

The version I have submitted to the school is the correct and final version, it is my own

work, and I have correctly acknowledged the work of others. I understand that failure to

do this will be investigated as a breach of IB regulations.

ALL RIGHTS BELONG TO OWNERTAKEN FROM WWW.INTERNALASSESSMENTS.WORDPRESS.COM

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Contents

Rationale ......................................................................................................................... 3

Introduction ..................................................................................................................... 3

Context of the exploration ......................................................................................... 3

Definitions of key terms ............................................................................................ 4

Main body ........................................................................................................................ 4

Integrating trigonometric functions with conventional methods ................................. 4

Proving Euler’s formula ............................................................................................. 7

Integrating trigonometric functions using Euler’s formula ......................................... 9

Trigonometric identities ........................................................................................... 13

Conclusion .................................................................................................................... 14

Bibliography .................................................................................................................. 15


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Rationale

In the Higher-Level Mathematics syllabus, we are required to know how to integrate

trigonometric functions. However, we do this without considering other more difficult

situations in which more than one method may not only be used, but is sometimes

imperative and a more productive approach.

As such, after doing research on integrating trigonometric functions, I discovered that

using Euler’s formula is a pivotal aspect in this area of mathematics. I have briefly

glanced the formula before and wanted to find out more about its relation to

trigonometric functions.

This was an unfamiliar experience because I had never been introduced to the concept

that bridges natural exponents and trigonometry, nor had I been introduced to imaginary

numbers.

Since this captured my personal interest and heightened my awareness in more areas

of mathematics, I decided to use this as my choice of topic for this exploration.

This exploration aims to find out the techniques of integrating trigonometric functions

and explain each of these techniques. The main question is, which method works best

in each situation? I will conclude with an evaluation of these methods in each different

situation.

Introduction

Context of the exploration

In the first year of the IB Diploma, my Grade 11 mathematics teacher showed me a

mathematical truth on the board for myself and my classmates to do some self-reading

on. Upon seeing it, I was stunned by its apparent simplicity and mathematical beauty.

The mathematical truth was: 𝒆𝒊𝝅 + 𝟏 = 𝟎

I was so shocked that 3 very important mathematical units would be so easily united

under a single, non-complex or irrational answer in such an arrangement. After doing

some research, I’ve discovered that this truth is derived from Euler’s formula, which is

written as such: 𝒆𝒊𝒙 = 𝐜𝐨𝐬 𝒙 + 𝒊 𝐬𝐢𝐧 𝒙 (Tsumura, 2004).

This allowed me to go further into the imaginary unit, which has yet to be covered in the

HL Mathematics syllabus, making this is for my own personal interest into the

exploration of unfamiliar mathematics, amongst other things covered in this exploration,

which is mostly outside of the IB HL Mathematics syllabus. It is also worth noting that

trigonometry and integral calculus are essential syllabus components, even though the

mathematics used in here often requires an alternative approach.


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Therefore, I chose it to be the subject of my exploration: not only to seek an experience

out of passion, but also to do further research on existing coursework and to develop

what I had learned from teachers by myself, effectively killing two birds with one stone.

Definitions of key terms

𝑖 – Known as the imaginary unit, used to define imaginary numbers which arises from

answers with imaginary solutions, like the square of a negative number or the non-real

root of quadratic functions. Its quantifiable value is √−1.

𝑒 – Known as Euler’s number, it is the asymptote of the converging function

(1 +1

𝑛)

𝑛

- it is the base of the natural exponential function, represented as 𝑒𝑥. Its

quantifiable value, to 3 decimal places, is 2.718.

𝜋 – The ratio between a circle’s diameter and circumference, also used in radian

trigonometry and other circular-based theorems, such as spirals and spheres. Its

quantifiable value is, to 3 decimal places, 3.142.

Taylor series – A Taylor series is a series expansion of a function about a point. A one-

dimensional Taylor series is an expansion of a real function 𝑓(𝑥) about a point 𝑥 = 𝑎 is

given by 𝑓(𝑎) + 𝑓′(𝑎)(𝑥 − 𝑎) +𝑓′′(𝑎)

2!(𝑥 − 𝑎)2 +

𝑓(3)(𝑎)

3!(𝑥 − 𝑎) 3 + ⋯ +

𝑓(𝑛)(𝑎)

𝑛!(𝑥 − 𝑎)𝑛 + ⋯

which expands infinitely (Weisstein, 2004). When 𝑎 = 0, it becomes known as a

Maclaurin series.

Main body

Integrating trigonometric functions with conventional methods

I will solve two functions: 𝑓(𝑥) = ∫ 𝑒−𝑥 cos 𝑥 𝑑𝑥 and 𝑓(𝑥) = ∫ sin6 𝑥 𝑑𝑥 with various

techniques.

Solving the first function using integration by parts

Finding the solution using the integration by parts formula will separate the whole

function into something that is easier to work with, even though this is a longer and

more tedious method since the integration is cumbersome and repetitive.

𝑓(𝑥) = ∫ 𝑒−𝑥 cos 𝑥 𝑑𝑥

= 𝑒−𝑥 sin 𝑥 − ∫(−𝑒−𝑥) (sin 𝑥) 𝑑𝑥

Since this presents another integral that cannot be worked on directly, integration by

parts must once again be used.


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𝑒−𝑥 sin 𝑥 − ∫(−𝑒−𝑥) (sin 𝑥) 𝑑𝑥 = 𝑒−𝑥 sin 𝑥 − 𝑒−𝑥 cos 𝑥 − ∫(−𝑒−𝑥) (−cos 𝑥) 𝑑𝑥

= 𝑒−𝑥 sin 𝑥 − 𝑒−𝑥 cos 𝑥 − ∫(𝑒−𝑥) (cos 𝑥) 𝑑𝑥

Since this has led back to the function to be solved itself appearing in the expansion, I

can substitute it in and solve it from there.

𝑓(𝑥) = 𝑒−𝑥 sin 𝑥 − 𝑒−𝑥 cos 𝑥 − 𝑓(𝑥)

2𝑓(𝑥) = 𝑒−𝑥 sin 𝑥 − 𝑒−𝑥 cos 𝑥

2𝑓(𝑥) = 𝑒−𝑥(sin 𝑥 − cos 𝑥)

𝑓(𝑥) = 𝑒−𝑥 (sin 𝑥 − cos 𝑥

2)

This method is limiting since the integration by parts not only needs to be carried out

twice, but also because as someone solving the issue, I would not be able to

instinctively carry out something out that did not break down into solvable parts the first

time. Hence, another method of solving this is imperative.

Solving the second function using trigonometric identities

Due to the fact that function has such a large exponent, this needs to broke down into a

smaller exponential to utilise trigonometric identities for the further simplification and

derivation. However, the Pythagorean identities cannot be used since those still use

exponents on both sides of the formula, hence necessitating the use of one the double

angle identities, where the square is only present on one side of the identity.

𝑓(𝑥) = ∫ sin6 𝑥 𝑑𝑥 = ∫(sin2 𝑥)3 𝑑𝑥

From here, I have observe that sin2 𝑥 is present in the identity cos 2𝑥 = cos2 𝑥 − sin2 𝑥.

Since cos2 𝑥 = 1 − sin2 𝑥 (from Pythagorean identities), cos 2𝑥 = 1 − 2sin2 𝑥. Thus,

sin2 𝑥 can alternatively be represented as 1−cos 2𝑥

2.

∫(sin2 𝑥)3 𝑑𝑥 = ∫ (1 − cos 2𝑥

2)

3

𝑑𝑥

From here, the numerator inside the brackets can be expanded using the binomial

theorem.

∫ (1 − cos 2𝑥

2)

3

𝑑𝑥 = ∫1 − 3 cos 2𝑥 + 3 cos2 2𝑥 − cos3 2𝑥

8𝑑𝑥


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Since integrals are summations of a function’s heights given a certain range on the axis,

they may be treated as summations and split up accordingly. On top of that, any

constants may be put outside of the integral sign, since they don’t need to be summed.

=1

8∫ 𝑑𝑥 −

3

8∫ cos 2𝑥 𝑑𝑥 +

3

8∫ cos2 2𝑥 𝑑𝑥 −

1

8∫ cos3 2𝑥 𝑑𝑥

The biggest problem now is the cos2 2𝑥 and cos3 2𝑥 parts of the function, which are

impossible to integrate in their current form. To solve for ∫ cos2 2𝑥 𝑑𝑥, use the double

angle identity again, equate to ∫cos 4𝑥+1

2𝑑𝑥.

To solve for ∫ cos3 2𝑥 𝑑𝑥, substitution is required, not just to reduce the value of the

exponential, but also make the integration avoid the trigonometric element, which would

result in it being easier to solve for.

Take 𝑦 = sin 2𝑥, 𝑑𝑦

𝑑𝑥= 2cos 2𝑥,

𝑑𝑦

2cos 2𝑥= 𝑑𝑥. Substitute 𝑑𝑥 into original function.

∫ cos3 2𝑥𝑑𝑦

2cos 2𝑥=

1

2∫ cos2 2𝑥 𝑑𝑦

Using the Pythagorean identities, cos2 2𝑥 = 1 − sin2 2𝑥 = 1 − 𝑦2. Thus,

1

2∫ cos2 2𝑥 𝑑𝑦 =

1

2∫ 1 − sin2 2𝑥 𝑑𝑦 =

1

2∫ 1 − 𝑦2 𝑑𝑦

This integral can now be worked on as a normal algebraic integration, due to the new

substitution of 𝑦 and integration with respect to 𝑦, then converted back into 𝑥 values.

1

2∫ 1 − 𝑦2 𝑑𝑦 =

1

2(𝑦 −

𝑦3

3) =

1

2(sin 2𝑥 −

sin3 2𝑥

3)

From here, solve for the rest of the function.

𝑓(𝑥) =1

8∫ 𝑑𝑥 −

3

8∫ cos 2𝑥 𝑑𝑥 +

3

8∫

cos 4𝑥 + 1

2𝑑𝑥 −

1

8×

1

2(sin 2𝑥 −

sin3 2𝑥

3)

=1

8𝑥 −

3

8× (

sin 2𝑥

2) +

3

8× (

sin 4𝑥

8+

1

2𝑥) −

1

16(sin 2𝑥 −

sin3 2𝑥

3)

=𝑥

8−

3 sin 2𝑥

16+

3 sin 4𝑥

64+

3𝑥

16−

sin 2𝑥

16+

sin3 2𝑥

48

=5𝑥

16−

sin 2𝑥

4+

3 sin 4𝑥

64+

sin3 2𝑥

48

Here, I have noted that this method required multiple integrals to be split and

substituted, which is a time-consuming method of solving for these integrals since they


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have such large exponents. Thus, a proof where this would not be a problem is

important, hence demanding the need for Euler’s formula.

Proving Euler’s formula

As stated before, certain functions in mathematics, including trigonometric functions,

may be expanded using the Taylor series or Maclaurin series. How they work is by

taking the same 𝑥-value on the graph (Maclaurin series takes 0), and then taking the

sum of all the differentials of that point, with each entity divided by the factorial of the

number of differentials, which will equate to the function itself, which leads to easily re-

representing the function as a series.

When 𝑠𝑖𝑛𝑥 and 𝑐𝑜𝑠𝑥 are expanded using the Maclaurin series, they yield a series of

whole numbers tied to 𝑥. The reason this is necessary is because it is used to compare

to other Maclaurin series to find any similarities to correlate them.

Keep in mind that while 𝑥 = 𝑎 = 0, 𝑠𝑖𝑛0 = 0, I still needed to stretch it out with respect to

𝑥 to make sense out of the expansion and thereby derive the formula.

When 𝑓(𝑥) = sin 𝑥 is expanded infinitely using the Maclaurin series, it is derived like

this:

sin 𝑥 = 𝑓(0) + 𝑓′(0)(𝑥 − 0) +𝑓′′(0)

2!(𝑥 − 0)2 +

𝑓(3)(0)

3!(𝑥 − 0)3 + +

𝑓(4)(0)

4!(𝑥 − 0)4 ⋯

= sin 0 + cos 0 ×𝑥 +(− sin 0)

2!×𝑥2 +

(− cos 0)

3!×𝑥3 +

sin 0

4!×𝑥4 …

= 0 + 1×𝑥 +(0)

2!×𝑥2 +

(−1)

3!×𝑥3 +

0

4!×𝑥4 …

Thus, can be written as:

sin 𝑥 = 𝑥 −𝑥3

3!+

𝑥5

5!−

𝑥7

7!+ ⋯

What I noticed is that the differential follows an alternating pattern of

sin 𝑎 , cos 𝑎 , − sin 𝑎 , − cos 𝑎 , sin 𝑎 … and since 𝑎 = 0, the pattern is 0,1,0, −1,0 …, while

still leaving the 𝑥𝑛 values behind.

Since cos 𝑥 also follows trigonometric differentiation, it will follow a pattern of its own too,

which would be cos 𝑎 , − sin 𝑎 , − cos 𝑎 , sin 𝑎 , cos 𝑎 …, and when 𝑎 = 0, the pattern is

1,0, −1,0,1 …

Likewise, when cos 𝑥 is expanded it is written like this:

cos 𝑥 = 1 −𝑥2

2!+

𝑥4

4!−

𝑥6

6!+ ⋯


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The numbers present in these expansions are not just the odd and even integers from

0, they are also present in the same manner of the next expansion, the natural

exponential function.

The natural exponential function is the function 𝑓(𝑥) = 𝑏𝑥, where 𝑎 is any real number.

Since the derivative is worked out for the above function is 𝑓’(𝑥) = 𝑏𝑥 log𝑒 𝑏, it means

that if 𝑏 = 𝑒, then it would become 𝑓(𝑥) = 𝑓’(𝑥).

Since differentiating this function would cause it to yield itself, the Maclaurin series,

which revolves upon differentiation, for 𝑓(𝑥) = 𝑒𝑥 would be:

𝑓(𝑥) = 𝑓(0) (1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯ ), since 𝑓(0) = 𝑓′(0) = 𝑓(𝑛)(0)

Because 𝑓(0) = 𝑒0 = 1, I simply get:

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯

Since the similarity to another Maclaurin series has been found, the intuitive approach

from here would be to take sin 𝑥 and cos 𝑥 and sum them up to get 𝑒𝑥, but since there

are negatives in the trigonometric expansions (unlike 𝑒𝑥), a way must be found to

equate these negative terms.

sin 𝑥 + cos 𝑥 ≠ 𝑒𝑥, but must be something along these lines.

Hence, this calls for the introduction of the role of the 𝑖 when attempting to sum these

trigonometric expansions attempt to get an answer in terms of 𝑒𝑥. Since 𝑖 = √−1, using

powers on it causes a repeating pattern, similarly to 1 or -1. It also helps that the pattern

involves alternating operations (+ and −). The pattern is:

𝑖 = √−1, 𝑖2 = −1, 𝑖3 = −𝑖, 𝑖4 = 1, 𝑖5 = 𝑖 and vice versa.

Firstly, I observed that in this pattern that if 𝑖 has an even exponent, the final result

leaves no 𝑖 behind. Hence, if I substitute the 𝑥 values in 𝑒𝑥 with 𝑖𝑥, then the result would

be that the terms with even exponents having no 𝑖’s. Since all the even-exponent terms

are simply extracted from cos 𝑥, I know that that part of the sum is left untouched.

Since the Maclaurin series is based on a function, if 𝑓(𝑥) = 𝑒𝑥, substituting the 𝑥 values

with 𝑖𝑥 would be 𝑓(𝑖𝑥), which is 𝑒𝑖𝑥. This technically works out, since the MacLaurin

series requires it to be represented as a function first before being expanded.

If 𝑓(𝑥) = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯, substitute 𝑥 for 𝑖𝑥 to obtain 𝑓(𝑖𝑥).

𝑓(𝑖𝑥) = 1 + 𝑖𝑥 +(𝑖𝑥)2

2!+

(𝑖𝑥)3

3!+ ⋯


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This can therefore be simplified into:

𝑒𝑖𝑥 = 1 + 𝑖𝑥 −𝑥2

2!+

𝑖𝑥3

3!+ ⋯

From here, I can remove the cos 𝑥 entities to leave behind the sin 𝑥 entities.

𝑒𝑖𝑥 − cos 𝑥 = 𝑖𝑥 −𝑖𝑥3

3!+

𝑖𝑥5

5!−

𝑖𝑥7

7!+ ⋯

With 𝑖 being a common multiple, I noted that:

𝑒𝑖𝑥 − cos 𝑥

𝑖= 𝑥 −

𝑥3

3!+

𝑥5

5!−

𝑥7

7!+ ⋯ = sin 𝑥

𝑒𝑖𝑥 − cos 𝑥

𝑖= sin 𝑥

𝑒𝑖𝑥 − cos 𝑥 = 𝑖 sin 𝑥

𝑒𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥

Hence obtaining 𝐜𝐨𝐬 𝒙 + 𝒊 𝐬𝐢𝐧 𝒙 = 𝒆𝒊𝒙.

In the unique case where 𝑥 = 𝜋, cos 𝜋 + 𝑖 sin 𝜋 = 𝑒𝑖𝜋. As mentioned before, 𝜋 is used in

trigonometry, and is the value of sin 𝑥 where it completes precisely half of its full period,

since trigonometric functions are periodic.

Because cos 𝜋 = −1 and 𝑖 sin 𝜋 = 0𝑖, the function above reduces to −1 + 0𝑖 = 𝑒𝑖𝜋,

−1 + 0 = 𝑒𝑖𝜋

−1 = 𝑒𝑖𝜋

0 = 𝑒𝑖𝜋 + 1

Obtaining 𝒆𝒊𝝅 + 𝟏 = 𝟎. This is known as Euler’s identity, which is largely considered

“The most beautiful function in mathematics” (Coolman, 2015).

Integrating trigonometric functions using Euler’s formula

Once again, I will solve 𝑓(𝑥) = ∫ 𝑒−𝑥 cos 𝑥 𝑑𝑥 and 𝑓(𝑥) = ∫ sin6 𝑥 𝑑𝑥.

The methods used to solve these functions with Euler’s formula are not present in the IB

mathematics syllabus or formula booklet for integration, so I had to personally work

these proofs out using the formula and my current understanding, adding to the

originality of this exploration.


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Solving the first function

Instead of using integration by parts, I will use Euler’s formula to substitute values into

the function instead, thus I will find identities for cos 𝑥 and sin 𝑥 in terms of natural

exponents.

𝑔(𝑥) = cos 𝑥 + 𝑖 sin 𝑥 = 𝑒𝑖𝑥, and likewise 𝑔(−𝑥) = cos −𝑥 + 𝑖 sin −𝑥 = 𝑒−𝑖𝑥

𝑔(−𝑥) = cos 𝑥 − 𝑖 sin 𝑥 = 𝑒−𝑖𝑥.

To cancel out the 𝑖 sin 𝑥 to isolate cos 𝑥, add 𝑔(𝑥) and 𝑔(−𝑥) to get

𝑒𝑖𝑥 + 𝑒−𝑖𝑥 = 2 cos 𝑥 , cos 𝑥 =𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2

The same can be done for sin 𝑥 by subtracting 𝑔(𝑥) and 𝑔(−𝑥) to get

𝑒𝑖𝑥 − 𝑒−𝑖𝑥 = 2𝑖 sin 𝑥 , sin 𝑥 =𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖

After that, simply substitute the identities into 𝑓(𝑥) to solve.

𝑓(𝑥) = ∫ 𝑒−𝑥 cos 𝑥 𝑑𝑥 = ∫ 𝑒−𝑥 (𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2) 𝑑𝑥 =

1

2∫ 𝑒−𝑥 (𝑒𝑖𝑥 + 𝑒−𝑖𝑥) 𝑑𝑥

Since I have arrived at 𝑒𝑖𝑥 and 𝑒−𝑖𝑥, I may once again utilise Euler’s formula into this

function to convert it back into trigonometry.

1

2∫ 𝑒−𝑥 (𝑒𝑖𝑥 + 𝑒−𝑖𝑥) 𝑑𝑥 =

1

2∫ 𝑒−𝑥+𝑖𝑥 + 𝑒−𝑥−𝑖𝑥 𝑑𝑥

The integral part can be simply converted since it involves single entities involving

purely natural exponents and no trigonometry added to it.

1

2∫ 𝑒−𝑥+𝑖𝑥 + 𝑒−𝑥−𝑖𝑥 𝑑𝑥 =

1

2(

𝑒𝑥(𝑖−1)

𝑖 − 1+

𝑒𝑥(−𝑖−1)

−𝑖 − 1)

Hence, this method only involves a single integration, which even so does not contain a

multiple conversions or repetitions that the integrations by parts method does, since the

integration involved in this method is a straightforward one involving just a slightly

different presentation of the natural exponential function (which, as discussed earlier,

will yield a differential or integral equal to the function).

From here, all I need to do is convert the denominators into the same thing to add up

the entities and then solve the function.

1

2(

𝑒𝑥(𝑖−1)

𝑖 − 1+

𝑒𝑥(−𝑖−1)

−𝑖 − 1) =

𝑒𝑥(𝑖−1)(−𝑖 − 1) + 𝑒𝑥(−𝑖−1)(𝑖 − 1)

2(𝑖 − 1)(−𝑖 − 1)


11

=−𝑖𝑒𝑥(𝑖−1) − 𝑒𝑥(𝑖−1) + 𝑖𝑒𝑥(−𝑖−1) − 𝑒𝑥(−𝑖−1)

2(−𝑖2 − 𝑖 + 𝑖 + 1)

= 𝑒−𝑥−𝑖𝑒𝑖𝑥 − 𝑒𝑖𝑥 + 𝑖𝑒−𝑖𝑥 − 𝑒−𝑖𝑥

4

= 𝑒−𝑥𝑖(−𝑒𝑖𝑥 + 𝑒−𝑖𝑥) − 𝑒𝑖𝑥 − 𝑒−𝑖𝑥

4

=𝑒−𝑥

2(

𝑖(−𝑒𝑖𝑥 + 𝑒−𝑖𝑥)

2−

𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2)

=𝑒−𝑥

2(

𝑖2(−𝑒𝑖𝑥 + 𝑒−𝑖𝑥)

2𝑖−


2) =

𝑒−𝑥

2(

(𝑒𝑖𝑥 − 𝑒−𝑖𝑥)

2𝑖−


2)

= 𝑒−𝑥 (sin 𝑥 − cos 𝑥

2)

Not only has this method allowed the solution to be found using a single integration, it

has also helped to generate more trigonometric identities in terms of 𝑒 and 𝑖, due to

their derivation via Euler’s formula.

However, this method might also be perceived as a limitation, because although the

mathematics used was far more straightforward and basic, it also became more time

consuming to do so thus.

Solving the second function

As proven earlier, sin 𝑥 =𝑒𝑖𝑥−𝑒−𝑖𝑥

2𝑖. Substitute into the second function.

𝑓(𝑥) = ∫ sin6 𝑥 𝑑𝑥 = ∫(sin 𝑥)6 𝑑𝑥 = ∫ (𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖)

6

𝑑𝑥

From here, use the binomial theorem again to expand the numerator. To make the

expansion easier to work with, substitute 𝑒𝑖𝑥 = 𝑦.

∫ (𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖)

6

𝑑𝑥 = ∫(𝑦 − 𝑦−1)6

64𝑖6𝑑𝑥

= −1

64∫ 𝑦6 − 6𝑦5𝑦−1 + 15𝑦4𝑦−2 − 20𝑦3𝑦−3 + 15𝑦2𝑦−4 − 6𝑦1𝑦−5 + 𝑦−6 𝑑𝑥

= −1

64∫ 𝑦6 − 6𝑦4 + 15𝑦2 − 20 + 15𝑦−2 − 6𝑦−4 + 𝑦−6 𝑑𝑥


12

Now, I may either resubstitute the value of 𝑒𝑖𝑥 = 𝑦, or integrate the whole function with

respect to 𝑦 instead by substituting for 𝑑𝑥. Either way is equally effective, but here I will

demonstrate the latter.

𝑦 = 𝑒𝑖𝑥,𝑑𝑦

𝑑𝑥= 𝑖𝑒𝑖𝑥, 𝑑𝑥 =

𝑑𝑦

𝑖𝑒𝑖𝑥=

𝑑𝑦

𝑖𝑦. With this information, the function becomes:

−1

64∫ 𝑦6 − 6𝑦4 + 15𝑦2 − 20 + 15𝑦−2 − 6𝑦−4 + 𝑦−6

𝑑𝑦

𝑖𝑦

= −1

64𝑖∫ 𝑦5 − 6𝑦3 + 15𝑦 − 20𝑦−1 + 15𝑦−3 − 6𝑦−5 + 𝑦−7 𝑑𝑦

= −1

64𝑖(

𝑦6

6− 6×

𝑦4

4+ 15×

𝑦2

2− 20 ln 𝑦 + 15×

𝑦−2

−2− 6×

𝑦−4

−4+

𝑦−6

−6)

= −𝑦6

384𝑖+

3𝑦4

128𝑖−

15𝑦2

128𝑖+

5 ln 𝑦

16𝑖+

15𝑦−2

128𝑖−

3𝑦−4

128𝑖+

𝑦−6

384𝑖

= − (𝑦6

384𝑖−

𝑦−6

384𝑖) + (

3𝑦4

128𝑖−

3𝑦−4

128𝑖) − (

15𝑦2

128𝑖−

15𝑦−2

128𝑖) +

5 ln 𝑦

16𝑖

= − ((𝑦2 − 𝑦−2)3 + 3𝑦2 − 3𝑦−2

384𝑖) +

3

64(

𝑦4 − 𝑦−4

2𝑖) −

15

64(

𝑦2 − 𝑦−2

2𝑖) +

5 ln 𝑦

16𝑖

= −1

48(

(𝑦2 − 𝑦−2)3

23𝑖) −

3

192(

𝑦2 − 𝑦−2

2𝑖) +

3

64(

𝑦4 − 𝑦−4

2𝑖) −

15

64(

𝑦2 − 𝑦−2

2𝑖) +

5 ln 𝑦

16𝑖

Now, I may convert the function back into terms of 𝑥. It works out into trigonometry

because of the following proof. Referencing the page containing proof of sin 𝑥 in the

natural exponential form:

𝑔(𝑥) − 𝑔(−𝑥)

2𝑖= sin 𝑥 =

𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖

This therefore means:

𝑔(𝑛𝑥) − 𝑔(−𝑛𝑥)

2𝑖=

(𝑒𝑖𝑥)𝑛

− (𝑒−𝑖𝑥)𝑛

2𝑖= sin 𝑛𝑥

Using this proof, I can finish solving the function when I resubstitute 𝑦 as 𝑒𝑖𝑥.

=1

48(

((𝑒𝑖𝑥)2 − (𝑒𝑖𝑥)−2)3

(2𝑖)3) −

3

192(

(𝑒𝑖𝑥)2 − (𝑒𝑖𝑥)−2

2𝑖) +

3

56(

(𝑒𝑖𝑥)4 − (𝑒𝑖𝑥)−4

2𝑖)

−15

56(

(𝑒𝑖𝑥)2 − (𝑒𝑖𝑥)−2

2𝑖) +

5 ln(𝑒𝑖𝑥)

16𝑖


13

=1

48sin3 2𝑥 −

3

192sin 2𝑥 +

3

64sin 4𝑥 −

15

64sin 2𝑥 +

5𝑥

16

=5𝑥

16−

sin 2𝑥

4+

3 sin 4𝑥

64+

sin3 2𝑥

48

Solving for the second function using Euler’s formula has not only been more hassle-

free in terms of the integration, since it only needs one integration to arrive at a final

answer. The last part was conducted simply to prove that the two methods of integration

are both mathematically valid and yield the same result.

Trigonometric identities

Beyond just integration, it is also worth noting that the trigonometric identities used to

solve the previous functions are also derivable using Euler’s formula, which I only

discovered this after some additional research and mathematics. This further amplifies

the importance and recognition Euler’s formula has in the field of integrating

trigonometric functions, because before solving the functions I must first know these

formulae.

Compound angle identities are the conversions or alternative representations of

sin 𝐴 ± 𝐵, cos 𝐴 ± 𝐵 and tan 𝐴 ± 𝐵. As such, to create identities for these, I substitute

these values in Euler’s formula.

cos 𝐴 ± 𝐵 + 𝑖 sin 𝐴 ± 𝐵 = 𝑒𝑖(𝐴±𝐵)

The properties of the natural exponentials can be exploited through this conversion,

hence making it possible to further break this down into identities.

= 𝑒𝑖(𝐴±𝐵) = 𝑒𝑖𝐴×𝑒±𝑖𝐵

This can once again be converted back into trigonometry using Euler’s formula.

(cos 𝐴 + 𝑖 sin 𝐴)(cos ±𝐵 + 𝑖 sin ±𝐵) = (cos 𝐴 + 𝑖 sin 𝐴)(± cos 𝐵 ± 𝑖 sin 𝐵)

This can be expanded and then separated into real and imaginary parts:

= ± cos 𝐴 cos 𝐵 ± 𝑖 cos 𝐴 sin 𝐵 ± 𝑖 sin 𝐴 cos 𝐵 ± 𝑖2 sin 𝐴 sin 𝐵

= ± cos 𝐴 cos 𝐵 ± 𝑖 cos 𝐴 sin 𝐵 ± 𝑖 sin 𝐴 cos 𝐵 ∓ sin 𝐴 sin 𝐵

= ± cos 𝐴 cos 𝐵 ∓ sin 𝐴 sin 𝐵 ± 𝑖(cos 𝐴 sin 𝐵 + sin 𝐴 cos 𝐵)

From here, I can equate the real and imaginary segments to the real and imaginary

segments of the original function.

cos 𝐴 ± 𝐵 = ± cos 𝐴 cos 𝐵 ∓ sin 𝐴 sin 𝐵 and sin 𝐴 ± 𝐵 = cos 𝐴 sin 𝐵 ∓ sin 𝐴 cos 𝐵

(International Baccalaureate Organisation, 2012).

Thus, I have acquired the compound angle identities for sin 𝐴 ± 𝐵 and cos 𝐴 ± 𝐵.


Trevor Lee

Sticky Note

Mistake :(

Trevor Lee

Sticky Note

It's not plus-minus, it's only positive.

14

To find the compound angle identity for tan 𝐴 ± 𝐵, simply convert it into sin 𝐴±𝐵

cos 𝐴±𝐵 and apply

the previously found compound angle identities to eventually acquire tan 𝐴±tan 𝐵

1∓tan 𝐴 tan 𝐵.

Next are the double angle identities, the other ways of expressing sin 2𝑥 , cos 2𝑥 and

tan 2𝑥. In a similar fashion to the deriving of the compound angle identities, I simply

substitute the trigonometric expressions into Euler’s formula.

cos 2𝑥 + 𝑖 sin 2𝑥 = 𝑒𝑖(2𝑥) = (𝑒𝑖𝑥)2

Since 𝑒𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥, the above formula may be rewritten as (cos 𝑥 + 𝑖 sin 𝑥)2,

which can then be expanded and once again separated into real and imaginary parts to

be compared to the original formula.

(cos 𝑥 + 𝑖 sin 𝑥)2 = cos2 𝑥 + 2𝑖 sin 𝑥 cos 𝑥 + 𝑖2 sin2 𝑥

= cos2 𝑥 + 2𝑖 sin 𝑥 cos 𝑥 − sin2 𝑥

= cos2 𝑥 − sin2 𝑥 + 𝑖(2 sin 𝑥 cos 𝑥)

Thus, cos 2𝑥 = cos2 𝑥 − sin2 𝑥 and sin 2𝑥 = 2 sin 𝑥 cos 𝑥 (International Baccalaureate

Organisation, 2012). Once again, to find the double angle identity for tan 2𝑥, convert it

into sin 2𝑥

cos 2𝑥 and utilise the previously found double angle formulas to arrive at

2 tan 𝑥

1−tan2 𝑥.

Another takeaway from this is that Euler’s formula acts as an effective bridge of natural

exponents and trigonometry, thus being able to convert and exploit properties from each

of these fields to come up with even more proofs, which would have been otherwise

impossible to do if Euler’s formula was not present, which emphasises the significance

of Euler’s formula in this exploration in relation to solving mathematics involving

trigonometry.

Conclusion

This exploration shown and explained the multiple techniques of integrating

trigonometric functions, with Euler’s formula playing a major role. To answer the

question proposed in the rationale, the solutions involving Euler’s formula for the

integrals were the most effective; while they did possess their limitations, the pros

outweigh the cons in terms of ease of integration, finding new identities with Euler’s

formula, and values for substitution were not as transparent as those used for the other

techniques.

However, this has not explored the full extent of the ability that can be provided by the

imaginary numbers, an important factor in both Euler’s formula and mathematics in

general, thus another exploration could be conducted with focus on said topic.

Euler’s formula has clear been shown as a breakthrough in mathematics as it allows

mathematicians to discover even more conversions. In this case, not only does it

involve imaginary numbers, it also combines trigonometry and natural exponentials.


15

However, there were various limitations with the exploration. For instance, there could

only be two functions used as examples of integrating trigonometric functions. Another

thing was that the answer was often kept in mind or known beforehand, especially for

the identities. For the identities, while the result did not need to be “known” as such, it

would not seem intuitively obvious to substitute the values into Euler’s formula to exploit

the exponential properties and converting them back into trigonometry. As for the extra

research done on this exploration, possible future investigations might include the

derivation, explanation and applications of Taylor and MacLaurin series, with proving

Euler’s formula being one such application.

This exploration has been an educational and insightful experience, not only have I

learnt a lot more about trigonometry and integration techniques, but I have also further

improved the immense knowledge of fundamentals that the IB HL Mathematics syllabus

demands, with rigorous proofs of more identities and creative approaches to substituting

values or new identities.

Bibliography

Coolman, R. (2015) Euler’s Identity: ‘The Most Beautiful Function’. Available at:

http://www.livescience.com/51399-eulers-identity.html (Accessed: 12 February 2017).

International Baccalaureate Organisation, (2012). Mathematics HL and further

mathematics HL formula booklet. 1st ed. Geneva: International Baccalaureate

Organisation, p.4.

Tsumura, H. (2004) ‘An elementary proof of Euler’s formula for z(2m)’, The American

Mathematical Monthly, 111(5), pp. 430–431. doi: 10.2307/4145270.

Weisstein, E.W. (2004) Taylor Series. Available at:

http://mathworld.wolfram.com/TaylorSeries.html (Accessed: 5 January 2017).


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