# Evaluate Integral: $$\cos x \cdot \cdot \cdot \cos 2^{2018}x$$

• MHB
• lfdahl
In summary, Opalg and I like Serena provided a thorough and clear explanation of the solution to the given integral problem. They also discussed a small typo in the solution and suggested an alternative solution. Both of them were thanked for their contributions and participation.
lfdahl
Gold Member
MHB
Evaluate

$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \: dx$

lfdahl said:
Evaluate

$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \: dx$

Is that -1 supposed to be there?

Prove It said:
Is that -1 supposed to be there?

Yes!

lfdahl said:
Evaluate

$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \: dx$

We can prove with induction and the product-to-sum-identity that:
$$\prod_{k=0}^n \cos 2^k x = \frac 1{2^n}\sum_{k=0}^{2^n-1}\cos (2k+1)x$$
Thus:
$$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \, dx \\ =\int_{0}^{2\pi}\left(\prod_{k=0}^{2017}\cos 2^kx\right) \cos (2^{2018}-1)x \, dx \\ =\frac 1{2^{2017}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\cos (2k+1)x \cos (2^{2018}-1)x \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi} 1 + \sum_{k=1}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \\ =\frac 1{2^{2018}}\left[ x + \sum_{k=1}^{2^{2018}-1}\frac{\sin(2^{k+1}-2)x}{2^{k+1}-2} \right]_{0}^{2\pi} \\ =\frac {2\pi}{2^{2018}} = \frac\pi{2^{2017}}\\$$

lfdahl said:
Evaluate

$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \: dx$
[sp]Using the identity $\cos\theta \cos\phi = \frac12\bigl(\cos(\theta+\phi) + \cos(\theta-\phi)\bigr)$, \begin{aligned} \cos x\cos 2x &= 2^{-1}(\cos x + \cos 3x) \\ \cos x\cos 2x\cos 4x &= 2^{-1}(\cos x + \cos 3x)\cos 4x = 2^{-2}(\cos x + \cos 3x + \cos 5x + \cos 7x) \\ \cos x\cos 2x\cos 4x \cos8x &= 2^{-3}(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos 15x) \\ &\vdots \\ \cos x\cos 2x\cos 4x \cdots \cos(2^nx) &= 2^{-n}\bigl(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos ((2^{n+1}-1)x)\bigr). \end{aligned} Then \begin{aligned}\cos x\cos 2x\cos 4x \cdots \cos(2^nx) \cos((2^{n+1}-1)x) &= 2^{-n}\bigl(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos ((2^{n+1}-1)x)\bigr)\cos((2^{n+1}-1)x) \\ &= 2^{-n-1}\bigl(1 + \cos 2x + \cos 4x + \cos 6x + \cos 8x + \ldots + \cos ((2^{n+2}-2)x)\bigr). \end{aligned} But $$\displaystyle \int_0^{2\pi}1\,dx = 2\pi,$$ and $$\displaystyle \int_0^{2\pi}\cos2k\pi\,dx = 0$$ for all nonzero integers $k$. Therefore $$\int_0^{2\pi} \cos x\cos 2x\cos 4x \cdots \cos(2^nx) \cos((2^{n+1}-1)x)\,dx = \frac{2\pi}{2^{n+1}} = \frac\pi{2^n}.$$ Finally, put $n=2017$ to get $$\int_0^{2\pi} \cos x\cos 2x\cos 4x \cdots \cos(2^{2017}x) \cos((2^{2018}-1)x)\,dx = \frac\pi{2^{2017}}.$$

[/sp]

Edit: Beaten to it by ILS!

I like Serena said:
We can prove with induction and the product-to-sum-identity that:
$$\prod_{k=0}^n \cos 2^k x = \frac 1{2^n}\sum_{k=0}^{2^n-1}\cos (2k+1)x$$
Thus:
$$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \, dx \\ =\int_{0}^{2\pi}\left(\prod_{k=0}^{2017}\cos 2^kx\right) \cos (2^{2018}-1)x \, dx \\ =\frac 1{2^{2017}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\cos (2k+1)x \cos (2^{2018}-1)x \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi} 1 + \sum_{k=1}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \\ =\frac 1{2^{2018}}\left[ x + \sum_{k=1}^{2^{2018}-1}\frac{\sin(2^{k+1}-2)x}{2^{k+1}-2} \right]_{0}^{2\pi} \\ =\frac {2\pi}{2^{2018}} = \frac\pi{2^{2017}}\\$$
Great job, I like Serena! Thankyou for your participation and a correct answer. Would you please help me understand the following step? I´m sure, you are right. It´s just my inert understanding of a detail in your nice solution:

It is the following equality:

$\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx$

Thankyou for being patient with me!(Blush)

Opalg said:
[sp]Using the identity $\cos\theta \cos\phi = \frac12\bigl(\cos(\theta+\phi) + \cos(\theta-\phi)\bigr)$, \begin{aligned} \cos x\cos 2x &= 2^{-1}(\cos x + \cos 3x) \\ \cos x\cos 2x\cos 4x &= 2^{-1}(\cos x + \cos 3x)\cos 4x = 2^{-2}(\cos x + \cos 3x + \cos 5x + \cos 7x) \\ \cos x\cos 2x\cos 4x \cos8x &= 2^{-3}(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos 15x) \\ &\vdots \\ \cos x\cos 2x\cos 4x \cdots \cos(2^nx) &= 2^{-n}\bigl(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos ((2^{n+1}-1)x)\bigr). \end{aligned} Then \begin{aligned}\cos x\cos 2x\cos 4x \cdots \cos(2^nx) \cos((2^{n+1}-1)x) &= 2^{-n}\bigl(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos ((2^{n+1}-1)x)\bigr)\cos((2^{n+1}-1)x) \\ &= 2^{-n-1}\bigl(1 + \cos 2x + \cos 4x + \cos 6x + \cos 8x + \ldots + \cos ((2^{n+2}-2)x)\bigr). \end{aligned} But $$\displaystyle \int_0^{2\pi}1\,dx = 2\pi,$$ and $$\displaystyle \int_0^{2\pi}\cos2k\pi\,dx = 0$$ for all nonzero integers $k$. Therefore $$\int_0^{2\pi} \cos x\cos 2x\cos 4x \cdots \cos(2^nx) \cos((2^{n+1}-1)x)\,dx = \frac{2\pi}{2^{n+1}} = \frac\pi{2^n}.$$ Finally, put $n=2017$ to get $$\int_0^{2\pi} \cos x\cos 2x\cos 4x \cdots \cos(2^{2017}x) \cos((2^{2018}-1)x)\,dx = \frac\pi{2^{2017}}.$$

[/sp]

Edit: Beaten to it by ILS!

Thankyou, Opalg for a very clear and lucid solution!

Just a small typo:

I´d expect the integrand $\cos 2k\pi$ to be $\cos 2kx$.

lfdahl said:
Great job, I like Serena! Thankyou for your participation and a correct answer. Would you please help me understand the following step? I´m sure, you are right. It´s just my inert understanding of a detail in your nice solution:

It is the following equality:

$\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx$

Thankyou for being patient with me!(Blush)

Sure!

It's a reordering of terms:
$\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \\ = \Big(\cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos \big(2^{2018}+2\cdot(2^{2017} -1)\big)x\Big) \\ \phantom{==} + \Big(\cos (2^{2018}-2)x + ... + \cos \big(2^{2018}-2\cdot(2^{2017}-1)-2\big)x\Big) \\ = \Big(\cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos (2^{2018+1} - 2)x\Big) \\ \phantom{==}+ \Big(\cos (2^{2018}-2)x + ... + \cos(2^{0+1}-2)x\Big)\\ = \cos(2^{0+1}-2)x + ... + \cos (2^{2018}-2)x + \cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos (2^{2018+1} - 2)x\\ =\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x$
(Angel)

I like Serena said:
Sure!

It's a reordering of terms:
$\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \\ = \Big(\cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos \big(2^{2018}+2\cdot(2^{2017} -1)\big)x\Big) \\ \phantom{==} + \Big(\cos (2^{2018}-2)x + ... + \cos \big(2^{2018}-2\cdot(2^{2017}-1)-2\big)x\Big) \\ = \Big(\cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos (2^{2018+1} - 2)x\Big) \\ \phantom{==}+ \Big(\cos (2^{2018}-2)x + ... + \cos(2^{0+1}-2)x\Big)\\ = \cos(2^{0+1}-2)x + ... + \cos (2^{2018}-2)x + \cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos (2^{2018+1} - 2)x\\ =\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x$
(Angel)

Hello again, I like Serena!

I´m still getting a different result, when I reorder the sum ... (Wondering) Please take a look:

To ease the algebra, I put $2^{2018}=N$.

This is the sum to be reordered:

$S = \sum_{k=0}^{\frac{N}{2}-1}\left ( \cos (N+2k)x + \cos (N-2k-2)x \right )$

The factors in the cosine arguments (for $k \in \left \{ 0,1,2,...,\frac{N}{2}-2, \frac{N}{2}-1\right \}$):

$(N+2k) \in \left \{ N,\, N+2, \, N+4, ... ,\, 2N-6,\, 2N-4, \, 2N-2 \right \}$
$(N-2k-2) \in \left \{ 0,\, 2, \, 4, ... ,\, N-6,\, N-4, \, N-2 \right \}$

This arithmetic progression is also obtained, when I look at the factor $2k$:
$2k \in \left \{ 0,2,4, ..., \, N-2,\, N,\, N+2, \, N+4, ...,2N-4, 2N-2\right \}$
for $k \in \left \{ 0,1,2, ... ,\, N-2, \, N-1 \right \}$

So our sum, $S$, reduces to:

$S = \sum_{k=0}^{N-1} \cos (2kx) = 1 + \sum_{k=1}^{2^{2018}-1} \cos (2kx)$

A small correction to I like Serena´s nice solution:

$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \, dx \\ =\int_{0}^{2\pi}\left(\prod_{k=0}^{2017}\cos 2^kx\right) \cos (2^{2018}-1)x \, dx \\ =\frac 1{2^{2017}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\cos (2k+1)x \cos (2^{2018}-1)x \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos 2kx \, dx \\ =\frac 1{2^{2018}} \left ( \int_{0}^{2\pi} 1 dx + \sum_{k=1}^{2^{2018}-1}\int_{0}^{2\pi}\cos 2kx \, dx \right ) \\ =\frac 1{2^{2018}}\left[ x + \sum_{k=1}^{2^{2018}-1} \frac{\sin 2kx}{2k} \right]_{0}^{2\pi} \\ =\frac {2\pi}{2^{2018}} = \frac\pi{2^{2017}}$

I have a suggested/alternative solution to the problem, but Opalg´s exemplary solution simply beats it.
Thankyou very much both of you, I like Serena and Opalg for your valuable contributions and participation!(Yes)

## What is the purpose of evaluating this integral?

The purpose of evaluating this integral is to find the exact value of the area under the curve of the given function. This can be used in various applications in physics, engineering, and other fields of science.

## What is the domain and range of the given function?

The domain of the function is all real numbers, while the range is between -1 and 1. This is because the cosine function oscillates between these two values.

## What are the steps to evaluate this integral?

The steps to evaluate this integral involve using trigonometric identities to simplify the expression and then using integration techniques such as u-substitution or integration by parts. The final step is to evaluate the integral using the fundamental theorem of calculus.

## Can this integral be solved without using trigonometric identities?

No, the use of trigonometric identities is necessary in order to simplify the expression and solve the integral. Without using these identities, the integral would be much more complex and difficult to solve.

## What is the significance of the exponent 2018 in the function?

The exponent 2018 represents the number of cosine functions that are being multiplied together. This means that the function is a composition of 2018 cosine functions, which can greatly affect the shape and behavior of the graph of the function.

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