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