how to calculuate $\int_0^ \pi \sqrt{1+x^2 \sin^2x}dx$

Question

I was finding arc length of $y=\sin x - x \cos x$ $(0 \leq x\leq \pi)$

and I found I've to solve

$$\int_0^\pi \sqrt{1 + x^2\sin^2{x}}\, dx $$

but I have no idea about this.

I tried using $\sin^2x + \cos^2x=1$, $\sin^2x =(1-\cos 2x)/2$ but failed.

Answer 1

Since there does not seem to be a simple closed form value, I would recommend the following simple approximate method.

Let the integrand $\sqrt{1 + x^2\sin^2{x}}$ have a maximum of M and minimum of m on $[0,\pi]$. Then the following inequality holds:

$$\pi \;\text{m}<I=\int_0^\pi \sqrt{1 + x^2\sin^2{x}}\, dx<\pi\;\text{M}$$

Minimum of the integrand is obviously $\text{m}=1$. Maximum of the integrand depends on the maximum of the function $f=(x\sin x)^2$. Below is a plot of $f$:

Since we are dealing with approximate calculation we take $f_{max}=3$ for sake of simplicity. (The actual value is $f_{max}=3.31...$). Thus, the maximum of the integrand is $\text{M}=2$ and we get the inequality:

$$\pi <I<2\pi$$

Now, as an estimation, we take the value of the integral to be the arithmetic mean of the upper and bottom bounds of the integral:

$$I=\frac{\pi+2\pi}{2}=\frac{3\pi}{2}=4.71...$$

The actual value of the integral is $I=4.69...$.

Of course, the fact that the estimation turned out to be so close to the actual value, is a fluke.