How can we prove $\int_1^\pi x \cos(\frac1{x}) dx<4$ by hand? Is there any way we can prove this definite integral inequality by hand:
$$
\int_{1}^{\pi}x\cos\left(1 \over x\right)\,{\rm d}x < 4
$$ 
I don't where to start even, please help. That
$\displaystyle\cos\left(1 \over x\right)\ \leq\ 1$ doesn't seem to help because $\displaystyle\int_{1}^{\pi}x\,{\rm d}x\ >\ 4$.
 A: Hint:
$$\begin{align}
∫_1^π x \cos \left(\frac1{x}\right)dx&=∫_{1}^{1/π}\frac{\cos {u}}{u}\left(-\frac{du}{u^2}\right)\\
&=∫_{1/π}^{1}\frac{\cos {u}}{u^3}du\\
&\leq∫_{1/π}^{1}\frac{1-\frac12u^2+\frac{1}{24}u^4}{u^3}du
\end{align}$$
A: First of all,
$$∫_{x=1}^π x\cos \frac{1}{x}dx=∫_{t=1}^\frac{1}{π} \frac{1}{t}*\cos \left(t\right)*\frac{-1}{t^2}dt$$
Next we have $$\cos \left(t\right) < 1 - \frac{t^2}{2}+\frac{t^4}{24}$$
Hence the given integral is smaller than
$$∫_{t=1}^\frac{1}{π} \frac{-1}{t^3}\left(1-\frac{t^2}{2}+\frac{t^4}{24}\right)dt=3.8811597.. $$
A: The integral of x dx is only a little bit greater than 4. 
For x < pi, cos (1/x) is not just ≤ 1. It's ≤ cos (1/pi). For x ≤ 2, it's ≤ cos (1/2). 
A: At the interval from $1/\pi$ to $1$ function $\cos(x)$ can be bounded from above by tangent line.
$$
\cos(x) \leq (\cos'(x)|_{\pi/6} ) (x - \frac{\pi}{6}) + \cos(\pi/6) = -\frac{1}{2} (x - \frac{\pi}{6}) + \cos(\frac{\pi}{6}).
$$
So the integral
$$
\int_1^\pi x \; \cos\left(\frac{1}{x}\right) \leq  \int_1^\pi  x \left( -\frac{1}{2} \left(\frac{1}{x} - \frac{\pi}{6}\right) + \cos\left(\frac{\pi}{6}\right)\right)
$$
the last integral can be easily computed by hand and equals $\approx 3.93...$
