Pythagorean identities will just have to do

Calculus Level 4

E = 0 π / 2 1 sin ( 2016 x ) d x \mathscr{E} = \displaystyle \int_{0}^{\pi/2} \sqrt{1- \sin(2016 x) }\, dx

Find E . \mathscr{E}.


Inspiration

Bonus: Generalize for 0 π / 2 1 sin ( 12 n x ) d x \displaystyle \int_{0}^{\pi/2} \sqrt{1- \sin(12n x) }\, dx , where n n is a positive integer , and prove it.


The answer is 1.414213562.

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2 solutions

Chew-Seong Cheong
Nov 12, 2017

E = 0 π 2 1 sin ( 4 k x ) d x Let θ = 4 k x d θ = 4 k d x = 1 4 k 0 2 k π 1 sin θ d θ Since the integrand has a period of 2 π = k 4 k 0 2 π 1 sin θ d θ See note. = 1 4 ( 0 π 1 sin θ d θ + π 2 π 1 sin θ d θ ) Since sin θ is symmetrical about π 2 and 3 π 2 = 1 4 ( 2 0 π 2 1 sin θ d θ + 2 π 3 π 2 1 sin θ d θ ) Using identity a b f ( x ) d x = a b f ( a + b x ) d x = 1 2 ( 0 π 2 1 sin ( π 2 θ ) d θ + π 3 π 2 1 sin ( π 2 θ ) d θ ) = 1 2 ( 0 π 2 1 cos θ d θ + π 3 π 2 1 cos θ d θ ) Since cos 2 u = 1 2 sin 2 u = 1 2 ( 0 π 2 2 sin θ 2 d θ + π 3 π 2 2 sin θ 2 d θ ) = 2 ( cos θ 2 π 2 0 + cos θ 2 3 π 2 π ) = 2 ( 1 1 2 + 0 + 1 2 ) = 2 \begin{aligned} \mathscr E & = \int_0^\frac \pi 2 \sqrt{1-\sin (4kx)} \ dx & \small \color{#3D99F6} \text{Let }\theta = 4kx \implies d \theta = 4k \ dx \\ & = \frac {\color{#3D99F6}1}{4k} \int_0^{\color{#3D99F6}2k\pi} \sqrt{1-\sin \theta} \ d\theta & \small \color{#3D99F6} \text{Since the integrand has a period of }2\pi \\ & =\frac {\color{#D61F06}k}{4k} \int_0^{\color{#D61F06}2\pi} \sqrt{1-\sin \theta} \ d\theta & \small \color{#D61F06} \text{See note.} \\ & = \frac 14 \left(\int_0^\pi \sqrt{1-\sin \theta} \ d\theta + \int_\pi^{2\pi} \sqrt{1-\sin \theta} \ d\theta \right) & \small \color{#3D99F6} \text{Since }\sin \theta \text{ is symmetrical about } \frac \pi 2 \text{ and } \frac {3\pi}2 \\ & = \frac 14 \left(2\int_0^\frac \pi 2 \sqrt{1-\color{#3D99F6}\sin \theta} \ d\theta + 2 \int_\pi^{\frac {3\pi}2} \sqrt{1-\color{#3D99F6}\sin \theta} \ d\theta \right) & \small \color{#3D99F6} \text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \left(\int_0^\frac \pi 2 \sqrt{1-\color{#3D99F6} \sin\left(\frac \pi 2- \theta\right)} \ d\theta + \int_\pi^{\frac {3\pi}2} \sqrt{1-\color{#3D99F6} \sin\left(\frac \pi 2- \theta\right)} \ d\theta \right) \\ & = \frac 12 \left(\int_0^\frac \pi 2 \sqrt{1-\color{#3D99F6} \cos \theta} \ d\theta + \int_\pi^{\frac {3\pi}2} \sqrt{1-\color{#3D99F6} \cos \theta} \ d\theta \right) & \small \color{#3D99F6} \text{Since }\cos 2u = 1-2\sin^2 u \\ & = \frac 12 \left(\int_0^\frac \pi 2 \sqrt 2 \sin \frac \theta 2 \ d\theta + \int_\pi^{\frac {3\pi}2} \sqrt 2 \sin \frac \theta 2 \ d\theta \right) \\ & = \sqrt 2 \left(\cos \frac \theta 2 \bigg|^0_\frac \pi 2 + \cos \frac \theta 2 \bigg|^\pi_{\frac {3\pi}2} \right) \\ & = \sqrt 2 \left(1 - \frac 1{\sqrt 2} + 0 + \frac 1{\sqrt 2} \right) \\ & = \sqrt 2 \end{aligned}

We note that the result applies to all multiples of 4, which 2016 is one. Therefore, 1000 E = 1414 \implies \lfloor 1000 \mathscr E\rfloor = \boxed{1414}


Note:

The figure is a plot of f ( x ) = 1 sin θ f(x) = \sqrt{1-\sin \theta} , we note that the area under the curve from 0 0 to 2 k π 2k\pi or k k cycles is the same as the area under the curve from 0 0 to 2 π 2\pi or 1 cycle multiplies by k k .

0 2 k π f ( x ) d x = k 0 2 π f ( x ) d x \implies \displaystyle \int_0^{2k\pi} f(x) \ dx = k \int_0^{2\pi} f(x) \ dx

Note another to find the integral of \rt(1-sinx) or is to multiply top and bottom by rt(1+sinx) which gives cosx/rt(1+sinx) which is an easy u sub to give 2rt(1+sinx)+C. Also works for rt(1-sinx)

D S - 3 years, 7 months ago

sqrt(1 - sin 2016x) = abs(sin 1008x - cos 1008x), then treat as 504 separate integrations, each giving (2 x sqrt2)/1008). The first quarter cycle (0 to pi/4032) and the last three-quarter one can be treated as one whole cycle

A Former Brilliant Member - 3 years, 6 months ago

Can you please explain how you got €=I/4

Navin Murarka - 3 years, 6 months ago

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I have changed the solution to explained it.

Chew-Seong Cheong - 3 years, 6 months ago

what if we make the integrand a perfect square an then simplify the two seperate integrals i.e (sin1008x-cos1008x)^2

It doesn' give the solutions even the integral calculator gives a different solution: CHECK THE FOLLOWING LINK BY PUTTING YOUR EXPRESSIONS...

https://www.symbolab.com/solver/definite-integral-calculator/%5Cint_%7B0%7D%5E%7B1.57%7D%5Csqrt%7B1-%5Csin%5Cleft(2016x%5Cright)%7Ddx

Ariijit Dey - 3 years, 6 months ago

I'm trying to understand step 3 (dropping 2 k π 2k\pi due to periodicity and having k pop out of integral) and step 6, where we get cos out instead of sin. Any help?

asd asdasd - 3 years, 6 months ago

Hello, first of all, I know I am wrong, but I am writing in order to somebody tells me where is my fail.

The primitive I achieve is 1008(Sqrt(1+sin(2016x)) and if we take the values x=0 and x=Pi/2, we obtain 0 as a result, which is absolutely false since the function is always positive. I know I am making a mistake, but I am not sure where...maybe in the process where cos(2016x) is involved?

Thank you for your help!!

Borja Cm - 3 years, 4 months ago
Boi (보이)
Nov 29, 2017

Lemme just

E = 0 π 2 1 sin ( 4 n x ) 1 + sin ( 4 n x ) 1 + sin ( 4 n x ) d x = 0 π 2 cos ( 4 n x ) 1 + sin ( 4 n x ) d x sin ( 4 n x ) is periodic over π 2 n . = n ( 0 π 8 n cos ( 4 n x ) 1 + sin ( 4 n x ) d x π 8 n 3 π 8 n cos ( 4 n x ) 1 + sin ( 4 n x ) d x + 3 π 8 n π 2 n cos ( 4 n x ) 1 + sin ( 4 n x ) d x ) t = sin ( 4 n x ) d t = 4 n cos ( 4 n x ) d x = n 1 4 n ( 0 1 1 1 + t d t 1 1 1 1 + t d t + 1 0 1 1 + t d t ) = 1 4 ( [ 2 1 + t ] 0 1 + [ 2 1 + t ] 1 1 + [ 2 1 + t ] 1 0 ) = [ 1 + t ] 1 1 = 2 . \mathscr E \\ \displaystyle = \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{1-\sin(4nx)}\sqrt{1+\sin(4nx)}}{\sqrt{1+\sin(4nx)}}dx \\ \displaystyle = \int_{0}^{\frac{\pi}{2}}\frac{|\cos(4nx)|}{\sqrt{1+\sin(4nx)}}dx ~~~~~\small \color{#3D99F6}\sin(4nx)\text{ is periodic over }\frac{\pi}{2n}. \\ \displaystyle = n\cdot\left(\int_{0}^{\frac{\pi}{8n}}\frac{\cos(4nx)}{\sqrt{1+\sin(4nx)}}dx - \int_{\frac{\pi}{8n}}^{\frac{3\pi}{8n}}\frac{\cos(4nx)}{\sqrt{1+\sin(4nx)}}dx + \int_{\frac{3\pi}{8n}}^{\frac{\pi}{2n}}\frac{\cos(4nx)}{\sqrt{1+\sin(4nx)}}dx \right) ~~~~~ \small \color{#3D99F6} t=\sin(4nx)~\Leftrightarrow~dt=4n\cos(4nx)dx \\ \displaystyle = n\cdot\frac{1}{4n}\left(\int_{0}^{1}\frac{1}{\sqrt{1+t}}dt - \int_{1}^{-1}\frac{1}{\sqrt{1+t}}dt + \int_{-1}^{0}\frac{1}{\sqrt{1+t}}dt\right) \\ \displaystyle = \frac{1}{4} \left(\left[2\sqrt{1+t}\right]_{0}^{1} + \left[2\sqrt{1+t}\right]_{-1}^{1} + \left[2\sqrt{1+t}\right]_{-1}^{0}\right) \\ \displaystyle =\left[\sqrt{1+t}\right]_{-1}^{1} \\ \displaystyle = \boxed{\sqrt{2}}.

You have to justify the first step.

Hobart Pao - 3 years, 6 months ago

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What do you mean?

Boi (보이) - 3 years, 6 months ago

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It's not clear how my original integral is equal to his/her first line. S/he skipped some important steps here.

Hobart Pao - 3 years, 6 months ago

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@Hobart Pao I don't see what's skipped.

I just multiplied 1 + sin ( 4 n x ) \sqrt{1+\sin(4nx)} to the numerator and denominator,

and used the identity 1 + sin ( 4 n x ) 1 sin ( 4 n x ) = 1 sin 2 ( 4 n x ) = cos 2 ( 4 n x ) = cos ( 4 n x ) . \sqrt{1+\sin(4nx)}\sqrt{1-\sin(4nx)}=\sqrt{1-\sin^2(4nx)}=\sqrt{\cos^2(4nx)}=|\cos(4nx)|.

Boi (보이) - 3 years, 6 months ago

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@Boi (보이) Nevermind, I'm stupid...sorry.

Hobart Pao - 3 years, 6 months ago

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@Hobart Pao It's okay!

And nah you're not, everyone makes mistakes ^^;

Boi (보이) - 3 years, 6 months ago

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