Isn't symmetry beautiful?

Calculus Level 2

$\displaystyle\int_{-2014}^{2014}\underbrace{\sin(\sin(\sin(\ldots\sin(\sin x)\ldots)))}_{\mbox{2014 times}}\, dx = \ ?$

Give your answer to 3 decimal places.

The answer is 0.

7 solutions

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Kenny Lau
Nov 27, 2014

informal version: since the function is odd, the area under the curve of the function from 0 to 2014 would be just the same as that from -2014 to 0, just 180 degrees rotated, so it would be 0.

formal version: $\begin{array}{clr} &\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} = -\underbrace{\sin(\sin(\sin(\cdots(\sin -x)\cdots)))}_{\mbox{2014 times}}\mbox{---(1)}\\ &\int_{0}^{2014}\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} dx\\ =&-\int_{2014}^{0}\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} dx\\ =&\int_{-2014}^{0}\underbrace{\sin(\sin(\sin(\cdots(\sin -u)\cdots)))}_{\mbox{2014 times}} du\\ =&-\int_{-2014}^{0}\underbrace{\sin(\sin(\sin(\cdots(\sin u)\cdots)))}_{\mbox{2014 times}} du\\ =&-\int_{-2014}^{0}\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} dx&\mbox{from (1)}\\ \therefore&\int_{-2014}^{2014}\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} dx\\ =&\int_{-2014}^{0}\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} dx + \int_{0}^{2014}\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} dx\\ =&\int_{-2014}^{0}\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} dx - \int_{-2014}^{0}\underbrace{\sin(\sin(\sin(\cdots(\sin x)\cdots)))}_{\mbox{2014 times}} dx\\ =&0 \end{array}$

Matthew Dickinson
Dec 2, 2014

Since the function is odd, the area under the curve of the function from 0 to 2014 would be just the same as that from -2014 to 0. Ergo; just 180 degrees but rotated, so it would be 0 primarily based on the fact that it's an odd function.

Sachin Arora
Dec 24, 2014

If we keep on taking sins of any no. The value keeps on getting less..I think if we would have integrated from 0 to 2014,answer would still be 0. Someone please very the same..

Moderator note:

This solution is wrong. We're not taking the sine functions infinitely many times, so the integrand is never $0$ when $x \ne 0 + \n \pi$ for positive integer $n$ .

Abhishek Kumar Sahu
Dec 2, 2014

Let the given expression = I Then use the properties of integral. F(x)=F(a+b-x), where a and b are limits of integral. Now sin(-x)=-sin(x) Hence 2I=0, hence I=0.

Bhargav Upadhyay
Dec 2, 2014

"sin()" is odd function. There are 2013 times sin() is there in side sin(), which ultimately turns in to an odd function. From basic property of finite integral answer equals to 0.

Vijay Hari Ram
Dec 4, 2014

leaving 2014 times, integration of sin is -cos x. substitutng values v will get 0.

Moderator note:

This solution is wrong. You can't have the integration $\sin [ f(x) ]$ of $-\cos [ f(x) ]$ . Differentiate and apply chain rule and you will see why it doesn't work.

Gulamo Jamal
Dec 2, 2014

cos⁡〖2014-cos⁡〖-2014 〗〗 =0.83-0.83=0

Moderator note:

The solution is wrong. The antiderivative of the integrand is not $\cos(x - \cos(x))$ , you can verify that by simply differentiate the expression with respect to $x$ .

It was not integrate sine x. but sine of sine of ... sine of x!

Kenny Lau - 6 years, 6 months ago

Isn't symmetry beautiful?

The answer is 0.

7 solutions

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