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1 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. 1

2 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 Conclusion Bibliography... 15

3 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: e iπ + 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: e ix = cos x + i sin x (Tsumura, 004). 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. 3

4 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 e x. Its quantifiable value, to 3 decimal places, is.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, Taylor series A Taylor series is a series expansion of a function about a point. A onedimensional 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)! (x a) + f(3) (a) 3! (x a) f(n) (a) (x a) n + n! which expands infinitely (Weisstein, 004). 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) = sin 6 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. 4

5 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) f(x) = e x sin x e x cos x f(x) = e x (sin x cos x) f(x) = e x sin x cos x ( ) 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) = sin 6 x dx = (sin x) 3 dx From here, I have observe that sin x is present in the identity cos x = cos x sin x. Since cos x = 1 sin x (from Pythagorean identities), cos x = 1 sin x. Thus, sin x can alternatively be represented as (sin x) 3 1 cos x dx = ( ) 3 dx 1 cos x From here, the numerator inside the brackets can be expanded using the binomial theorem. 1 cos x ( ) 3 dx = 1 3 cos x + 3 cos x cos 3 x dx 8. 5

6 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 x dx cos x dx 1 8 cos3 x dx The biggest problem now is the cos x and cos 3 x parts of the function, which are impossible to integrate in their current form. To solve for cos x dx, use the double angle identity again, equate to cos 4x+1 dx. To solve for cos 3 x 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 x, dy dy = cos x, dx cos x cos 3 dy x cos x = 1 cos x dy = dx. Substitute dx into original function. Using the Pythagorean identities, cos x = 1 sin x = 1 y. Thus, 1 cos x dy = 1 1 sin x dy = 1 1 y 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 1 y dy = 1 y3 (y 3 ) = 1 (sin x sin3 x ) 3 From here, solve for the rest of the function. f(x) = 1 8 dx 3 8 cos x dx cos 4x + 1 dx (sin x sin3 x ) 3 = 1 8 x 3 x (sin 8 ) + 3 4x (sin x) 1 16 (sin x sin3 x ) 3 = x 8 3 sin x sin 4x 64 = 5x sin x 3 sin 4x + + sin3 x x sin x sin3 x 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 6

7 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 rerepresenting 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)! = sin 0 + cos 0 x + ( sin 0) x +! (x 0) + f(3) (0) 3! ( cos 0) 3! = x + (0)! x + ( 1) x ! 4! x4 Thus, can be written as: sin x = x x3 3! + x5 5! x7 7! + (x 0) f(4) (0) (x 0) 4 4! x 3 + sin 0 x 4 4! 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 x n 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 x! + x4 4! x6 6! + 7

8 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) = b x, where a is any real number. Since the derivative is worked out for the above function is f (x) = b x log e 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) = e x would be: f(x) = f(0) (1 + x + x! + x3 3! + ), since f(0) = f (0) = f (n) (0) Because f(0) = e 0 = 1, I simply get: e x = 1 + x + x! + 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 e x, but since there are negatives in the trigonometric expansions (unlike e x ), a way must be found to equate these negative terms. sin x + cos x e x, 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 e x. 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, i = 1, i 3 = i, i 4 = 1, i 5 = 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 e x 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) = e x, substituting the x values with ix would be f(ix), which is e ix. 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 + x + x3 +, substitute x for ix to obtain f(ix).! 3! f(ix) = 1 + ix + (ix)! + (ix)3 3! + 8

9 This can therefore be simplified into: e ix = 1 + ix x! + ix3 3! + From here, I can remove the cos x entities to leave behind the sin x entities. e ix cos x = ix ix3 3! + ix5 5! ix7 7! + With i being a common multiple, I noted that: e ix cos x i e ix cos x i = x x3 3! + x5 5! x7 + = sin x 7! = sin x e ix cos x = i sin x e ix = cos x + i sin x Hence obtaining cos x + i sin x = e ix. In the unique case where x = π, cos π + i sin π = e iπ. 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 = e iπ, = e iπ 1 = e iπ 0 = e iπ + 1 Obtaining e iπ + 1 = 0. This is known as Euler s identity, which is largely considered The most beautiful function in mathematics (Coolman, 015). Integrating trigonometric functions using Euler s formula Once again, I will solve f(x) = e x cos x dx and f(x) = sin 6 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. 9

10 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 = e ix, 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 e ix + e ix = cos x, cos x = eix + e ix The same can be done for sin x by subtracting g(x) and g( x) to get e ix e ix = sin x, sin x = eix e ix After that, simply substitute the identities into f(x) to solve. f(x) = e x cos x dx = e x ( eix + e ix ) dx = 1 e x (e ix + e ix ) dx Since I have arrived at e ix and e ix, I may once again utilise Euler s formula into this function to convert it back into trigonometry. 1 e x (e ix + e ix ) dx = 1 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 e x+ix + e x ix dx = 1 (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 (ex(i 1) i 1 + ex( i 1) i 1 ) = ex(i 1) ( i 1) + e x( i 1) (i 1) (i 1)( i 1) 10

11 = iex(i 1) e x(i 1) + ie x( i 1) e x( i 1) ( i i + i + 1) = e x ieix e ix + ie ix e ix 4 = e x i( eix + e ix ) e ix e ix 4 = e x + e ix ) (i( eix = e x ( e ix + e ix ) (i = e x sin x cos x ( ) eix + e ix ) eix + e ix ) = e x e ix ) ((eix eix + e ix ) 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. Substitute into the second function. f(x) = sin 6 x dx = (sin x) 6 dx = ( eix e ix ) From here, use the binomial theorem again to expand the numerator. To make the expansion easier to work with, substitute e ix = y. ( eix e ix 6 ) dx = (y y 1 ) 6 64i 6 dx = 1 64 y6 6y 5 y y 4 y 0y 3 y y y 4 6y 1 y 5 + y 6 dx = 1 64 y6 6y y y 6y 4 + y 6 dx 6 dx 11

12 Now, I may either resubstitute the value of e ix = 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 = e ix, dy = dx ieix, dx = dy ie dy ix = iy. With this information, the function becomes: 1 64 y6 6y y y 6y 4 dy 6 + y iy = 1 64i y5 6y y 0y y 3 6y 5 + y 7 dy = 1 y4 y y y 4 64i (y ln y y 6 6 ) = y6 384i + 3y4 18i 15y 5 ln y y 18i 16i 18i 3y 4 18i + y 6 384i = ( y6 384i y 6 384i ) + ( 3y4 18i 3y 4 18i ) (15y 18i 15y 5 ln y ) + 18i 16i = ( (y y ) 3 + 3y 3y ) + 3 y 4 384i 64 (y4 ) 15 y 64 (y ) + 5 ln y 16i = 1 y ) 3 48 ((y 3 ) 3 y i 19 (y ) + 3 y 4 64 (y4 ) 15 y 64 (y ) + 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) = sin x = eix e ix This therefore means: g(nx) g( nx) = (eix ) n (e ix ) n = sin nx Using this proof, I can finish solving the function when I resubstitute y as e ix. = 1 ) (e ix ) 3 ) 48 (((eix () 3 ) 3 ) (e ix ) 19 ((eix ) + 3 ) 4 (e ix ) 4 56 ((eix ) 15 ) (e ix ) 56 ((eix ) + 5 ln(eix ) 16i 1

13 = 1 48 sin3 x x sin x + sin 4x sin x = 5x sin x 3 sin 4x + + sin3 x Solving for the second function using Euler s formula has not only been more hasslefree 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 = e i(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. = e i(a±b) = e ia 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 ± i 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, 01). Thus, I have acquired the compound angle identities for sin A ± B and cos A ± B. 13

14 To find the compound angle identity for tan A ± B, simply convert it into the previously found compound angle identities to eventually acquire sin A±B cos A±B tan A±tan B. 1 tan A tan B and apply Next are the double angle identities, the other ways of expressing sin x, cos x and tan x. In a similar fashion to the deriving of the compound angle identities, I simply substitute the trigonometric expressions into Euler s formula. cos x + i sin x = e i(x) = (e ix ) Since e ix = cos x + i sin x, the above formula may be rewritten as (cos x + i sin x), 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) = cos x + sin x cos x + i sin x = cos x + sin x cos x sin x = cos x sin x + i( sin x cos x) Thus, cos x = cos x sin x and sin x = sin x cos x (International Baccalaureate Organisation, 01). Once again, to find the double angle identity for tan x, convert it sin x tan x into and utilise the previously found double angle formulas to arrive at. cos x 1 tan 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. 14

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. (015) Euler s Identity: The Most Beautiful Function. Available at: (Accessed: 1 February 017). International Baccalaureate Organisation, (01). Mathematics HL and further mathematics HL formula booklet. 1st ed. Geneva: International Baccalaureate Organisation, p.4. Tsumura, H. (004) An elementary proof of Euler s formula for z(m), The American Mathematical Monthly, 111(5), pp doi: / Weisstein, E.W. (004) Taylor Series. Available at: (Accessed: 5 January 017). 15