Estimating $\int_{0}^{\pi}\sin(\sin(x))\,\mathrm{d}x$

Question

Which of the following is closest to this integral?

$$\int_{0}^{\pi}\sin(\sin(x))\,\mathrm{d}x$$

$$\text{(A) }1+\frac{\sqrt{2}}{2}\quad\quad\quad\quad\text{(B) }\sqrt{3}\quad\quad\quad\quad\text{(C) }\sqrt{\pi}\quad\quad\quad\quad\text{(D) }1+\frac{\pi}{4}\quad\quad\quad\quad\text{(E) }2\sqrt{2}-1$$

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This problem is to be answered in less than $4$ minutes, without a calculator.

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My Attempt: (Lengthy and consumes time (more than $4$ minutes)):

Recall that

$$\tag{1}\sin(x) = x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\dfrac{x^7}{5040}+\dots$$

Therefore, replacing $x$ with $\sin(x)$ results:

$$\tag{2}\sin(\sin(x))\approx \sin(x)-\dfrac{\sin^3(x)}{6}+\dfrac{\sin^5(x)}{120}-\dfrac{\sin^7(x)}{5040}$$

Also, recall that

$$\tag{3}\int_{0}^{\pi}\sin^{2k+1}(x)\,\mathrm{d}x=2\cdot\frac{2\cdot 4\cdot 6\cdot\dots\cdot (2k)}{3\cdot 5\cdot 7\cdot\dots\cdot (2k+1)}$$

Now, integrating $(2)$ term by term using $(3)$, we get:

$$\int_{0}^{\pi}\sin(\sin(x))\,\mathrm{d}x\approx \int_{0}^{\pi}\sin(x)\,\mathrm{d}x-\frac{1}{6}\int_{0}^{\pi}\sin^3(x)\,\mathrm{d}x+\frac{1}{120}\int_{0}^{\pi}\sin^5(x)\,\mathrm{d}x-\frac{1}{5040}\int_{0}^{\pi}\sin^7(x)\,\mathrm{d}x$$

$$\tag{4}=2-\frac{1}{6}\cdot\frac{4}{3}+\frac{1}{120}\cdot\frac{16}{15}-\frac{1}{5040}\cdot\frac{32}{35}\approx 1.786$$

Now, estimating the given options:

$\text{(A) }\approx 1+\dfrac{1.41}{2}=1.705$

$\text{(B) }\approx 1.73$

$\text{(C) }\approx \sqrt{3.14}\approx 1.77$ (with the help of Newton's Method)

$\text{(D) }\approx 1+\dfrac{3.14}{4}=1+0.785=1.785$

$\text{(E) }\approx 2\times 1.41-1=2.82-1=1.82$

Finally, $\text{(D)}$ seems the correct answer, but not confident enough since $\text{(C)}$ is also very close, and we are only estimating.

For verification only, I have used the calculator and conclude that $\text{(D)}$ is correct.

I am looking for a faster and more efficient way than this.

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Comments:

$(1)$ is the Taylor series expansion of $\sin(x)$

$(2)$ is the Taylor series expansion of $\sin(\sin(x))$

$(3)$ Wallis' Integral

$(4)$ Some tedious calculations

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Assumption:

We know the value of $\sqrt{2}$, $\sqrt{3}$, and $\pi$ for two decimal places.

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Your help would be appreciated. Thanks.

Answer 1

By the symmetry of $\sin \left(\sin x \right)$ about $\frac{\pi}{2}$, \begin{align*} I &= \int^{\pi}_{0} \sin \left(\sin x \right) dx \\ &= 2 \int^{\pi/2}_{0} \sin \left(\sin x \right) dx \end{align*}

Substituting $u=\sin x$, we get $\frac{du}{dx}=\cos x=\sqrt{1-u^2}$. So, $$I = 2 \int^{1}_{0} \frac{\sin t}{\sqrt{1-t^2}} dt$$

Using $\sin t \approx t-\frac{t^3}{6}+\frac{t^5}{120}$, we have that \begin{align*} I &= 2 \left(\int^{1}_{0} \frac{t}{\sqrt{1-t^2}} dt-\frac{1}{6} \int^{1}_{0} \frac{t^3}{\sqrt{1-t^2}} dt +\frac{1}{120} \int^{1}_{0} \frac{t^5}{\sqrt{1-t^2}} \right) \\ &= \frac{16}{9}+\frac{2}{225} \\ & \approx 1.7866 \end{align*}

The error from this approximation is bounded above by \begin{align*} 2 \cdot \frac{1}{7!} \int^{1}_{0} \frac{t^7}{\sqrt{1-t^2}} dt & < \frac{1}{2520} \int^{1}_{0} \frac{1}{\sqrt{1-t^2}} dt \\ &= \frac{\pi}{5040} \\ & < \frac{4}{5040} \\ &< 0.001 \end{align*} and so the answer is (D).

Answer 2

Just for the fun.

$$I=\int\frac{\sin (x)}{\sqrt{1-x^2}}\,dx$$ does not show any solution even using special functions.

For the pleasure of using once more my favored $1,400^+$ years old approximation of the since function $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad \text{for} \qquad 0\leq x\leq\pi$$ using factorization and partial fraction decomposition, we have $$\sin(x) \simeq -4 -\frac{4 a (a-\pi )}{(a-b) (x-a)}+\frac{4 b (b-\pi )}{(a-b) (x-b)}\qquad \text{where}\qquad (a,b)=\left(\frac{1}{2}\pm i\right) \pi$$

Using twice $$J(c)=\int \frac{dx}{(x-c)\sqrt{1-x^2} }=-\frac{2}{\sqrt{c^2-1}}\tan ^{-1}\left(\frac{x\sqrt{c^2-1}}{(x-c)+c \sqrt{1-x^2}}\right)$$ we have an explicit approximate antiderivative.

For your problem where we need $$A(c)=\int_0^1 \frac{dx}{(x-c)\sqrt{1-x^2} }$$ we have $$A(c)=\frac{2}{\sqrt{c^2-1}}\Bigg(\tan ^{-1}\left(\sqrt{c^2-1}\right)-\cot ^{-1}\left(\frac{1-c}{\sqrt{c^2-1}}\right) \Bigg)$$

Computing for a few values of $$K(t)=\int_0^t\frac{\sin (x)}{\sqrt{1-x^2}}\,dx$$ (the explicit result being converted to decimals) $$\left( \begin{array}{ccc} t & \text{approximation} & \text{numerical integration} \\ 0.00 & 0.000000 & 0.000000 \\ 0.05 & 0.001271 & 0.001251 \\ 0.10 & 0.005082 & 0.005008 \\ 0.15 & 0.011438 & 0.011293 \\ 0.20 & 0.020363 & 0.020137 \\ 0.25 & 0.031898 & 0.031588 \\ 0.30 & 0.046100 & 0.045714 \\ 0.35 & 0.063054 & 0.062600 \\ 0.40 & 0.082869 & 0.082362 \\ 0.45 & 0.105689 & 0.105145 \\ 0.50 & 0.131703 & 0.131141 \\ 0.55 & 0.161161 & 0.160599 \\ 0.60 & 0.194395 & 0.193853 \\ 0.65 & 0.231858 & 0.231356 \\ 0.70 & 0.274186 & 0.273741 \\ 0.75 & 0.322308 & 0.321940 \\ 0.80 & 0.377671 & 0.377398 \\ 0.85 & 0.442729 & 0.442571 \\ 0.90 & 0.522293 & 0.522273 \\ 0.95 & 0.628803 & 0.628962 \\ 1.00 & 0.892680 & 0.893244 \\ \end{array} \right)$$

So, for the integral, the approximation gives $1.78536$ which corresponds to an absolute error of $0.001128$ and a relative error of $0.063$%.

This is not doable in four minutes (even with a computer)

Edit 1

In the past, I worked approximations of the type $$\sin(x)=\sum_{n=1}^p a_n\, \frac {\big((\pi -x) x\big)^n } {\pi ^{2 n-1} }$$ where the first coefficients (easy to obtain using Taylor expansions) are $$\left( \begin{array}{cc} n & a_n \\ 1 & 1 \\ 2 & 1 \\ 3 & \frac{1}{6} \left(12-\pi ^2\right) \\ 4 & \frac{1}{2} \left(10-\pi ^2\right) \\ 5 & \frac{1}{120} \left(1680-180 \pi ^2+\pi ^4\right) \\ 6 & \frac{1}{24} \left(1008-112 \pi ^2+\pi ^4\right) \\ \end{array} \right)$$

We know that $$\int \frac{x^n}{\sqrt{1-x^2}}\,dx=\frac{x^{n+1}}{n+1}\,\,\, _2F_1\left(\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};x^2\right)$$

So, we have an explicit and simple approximation of the resuired antiderivative.

In particular for the present problem, we also know that $$\int_0^1 \frac{x^n}{\sqrt{1-x^2}}\,dx=\frac{\sqrt{\pi } \,\,\Gamma \left(\frac{n+1}{2}\right)}{2\, \Gamma \left(\frac{n+2}{2}\right)}$$

Converting the explicit result for the definite integral, we have $$\left( \begin{array}{cc} p & \text{integral} \\ 1 & 0.750000 \\ 2 & 0.883903 \\ 3 & 0.892913 \\ 4 & 0.893236 \\ 5 & 0.893244 \\ 6 & 0.893244 \\ \end{array} \right)$$

Three terms are more than sufficient for the result which is $$\frac{8316-3393 \pi ^2+2830 \pi^4}{2880 \pi ^4}$$

Edit 2

Much better would be to consider the norm $$\Phi=\int_0^\pi \Bigg( \sin(x)-\sum_{n=1}^2 a_n\, \frac {\big((\pi -x) x\big)^n } {\pi ^{2 n-1} }\Bigg)^2\,dx$$ which gives $$a_1=\frac{1680 \left(11 \pi ^2-108\right)}{\pi ^6} \qquad \qquad a_2=\frac{5040 \left(168-17 \pi ^2\right)}{\pi ^6}$$ which gives a maximum error of $8.887\times 10^{-4}$.

With this, the value of the considered definite integral is $$-\frac{105 \left(9240-1655 \pi ^2+72 \pi ^4\right)}{\pi ^8}=0.893476$$ to be compared to the exact $0.893244$.

The approximate antiderivative is, for this case, given by $$\frac{\pi ^9}{210}\,I=\left(3024+4590 \pi ^2-496 \pi ^4\right) \tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}-1}\right)+P(x)\,\sqrt{1-x^2}+C$$ where $$P(x)=\left(102 \pi ^2-1008\right) x^3+\left(2688 \pi -272 \pi ^3\right) x^2+\left(-1512-2295 \pi ^2+248 \pi ^4\right) x+\left(5376 \pi +320 \pi ^3-88 \pi ^5\right)$$

Over the whole range, the difference is less than $3\times 10^{-4}$.

Edit 3

Using AI Assist on the site, beside the exact $\frac{\pi }{2} \pmb{H}_0(1)$ already given in comments $$\int_0^1 \frac{\sin (x)}{\sqrt{1-x^2}}\, dx=2 \sum _{m=0}^{\infty } \frac{J_{2 m+1}(1)}{2 m+1}$$ Using two terms $$2 J_1(1)+\frac{2 J_3(1)}{3}=0.893143$$

Answer 3

An alternative would be to approximate the sine with a quadratic polynomial:

$$\sin x \approx \frac{2x}{\pi}+cx\left ( \frac{\pi}{2}-x \right )$$

where $$c=12\frac{4-\pi}{\pi^3}$$

Thus, we need to integrate the expression $$\frac{2\sin x}{\pi}+c\sin x\left ( \frac{\pi}{2}-\sin x \right )$$

from $0$ to $\frac{\pi}{2}$

and multiply the result by $2$:

$$I\approx 2\frac{12-\pi}{\pi^2} $$

Absolute error is about $0.0086$