Review question

# Can the integral of $\sin(\sin t)$ be zero? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6622

## Solution

In the region $0 < x \le 2\pi$, the equation $\int_0^x \sin(\sin t)\, dt = 0$ has

1. no solution;

2. one solution;

3. two solutions;

4. three solutions.

Can we evaluate this integral? It looks to be hard - impossible even! Instead, we’ll try to pick up some clues about the shape of $y = \sin(\sin t)$.

Clue 1. $\sin(\sin t)=0$ if and only if $\sin t = 0$, which happens just when $t = n\pi$, so that’s at $\pi$ and $2\pi$ on our interval (and at $0$).

Clue 2. $\sin t$ is positive for $0 < t < \pi$, taking values from $0$ to $1$ on the interval, so $\sin(\sin t)$ is also positive on this interval, taking values from $0$ to $\sin 1$.

Clue 3. $\sin t$ is negative for $\pi < t < 2\pi$, taking values from $0$ to $-1$ on this interval, so $\sin(\sin t)$ is also negative on this interval, taking values from $0$ to $-\sin 1$.

So the curve has a positive section from $0$ to $\pi$, and a negative section from $\pi$ to $2\pi$.

Clue 4. The function $y=\sin t$ is periodic; is $y = \sin(\sin t)$? We have $\sin(\sin(t+2\pi)) = \sin(\sin t)$, so yes, $\sin(\sin t)$ is periodic.

Could the period be less than $2\pi$? Clues 2. and 3. suggest not. We now know the section immediately on the other side of the origin is a negative section.

Clue 5. Could $\sin(\sin t)$ be odd or even? We have $\sin(\sin -t) = \sin(-\sin t) = -\sin(\sin t)$, and so $y=\sin(\sin t)$ is odd, and so has rotational symmetry about the origin.

This must mean that the negative section of the curve from $-\pi$ to $0$ matches exactly with the positive section from $0$ to $\pi$ (and the section from $\pi$ to $2\pi$, by periodicity).

We have enough now to answer the question; the (positive) area below the curve from 0 to $\pi$ matches exactly with the (negative) area under the curve from $\pi$ to $2\pi,$ so there is just one value of $x$ when the integral is zero ($x = 2\pi$) and the answer is (b).

So what does $y = \sin(\sin t)$ look like? The curve (in red below) is remarkably similar to the curve $y = \sin t$ (in blue below), but with a lesser amplitude.