Calculate the integral: integral from 1/pi to (2/pi)

Calculate the integral: $\int_{\frac{1}{\pi}}^{\frac{2}{\pi}} \frac{1}{x^2} \sin \frac{1}{x} \, dx$.

We will use substitution to simplify this integral.

Step 1: Choose a substitution

Let $u = \frac{1}{x}$. Then $x = \frac{1}{u}$ and $dx = -\frac{1}{u^2} \, du$. Also, $\frac{1}{x^2} = u^2$.

Step 2: Change the limits of integration

When $x = \frac{1}{\pi}$: $u = \pi$. When $x = \frac{2}{\pi}$: $u = \frac{\pi}{2}$.

Step 3: Rewrite the integral

\[\int_{\frac{1}{\pi}}^{\frac{2}{\pi}} \frac{1}{x^2} \sin \frac{1}{x} \, dx = \int_\pi^{\frac{\pi}{2}} u^2 \sin(u) \cdot \left(-\frac{1}{u^2}\right) du = -\int_\pi^{\frac{\pi}{2}} \sin u \, du\]

Step 4: Reverse the limits to remove the negative sign

\[= \int_{\frac{\pi}{2}}^\pi \sin u \, du\]

Step 5: Find the antiderivative

\[\int \sin u \, du = -\cos u + C\]

Step 6: Apply the Newton-Leibniz formula

\[\int_{\frac{\pi}{2}}^\pi \sin u \, du = \left[-\cos u\right]_{\frac{\pi}{2}}^\pi\]

Step 7: Evaluate at the limits

At $u=\pi$: $-\cos(\pi)=1$. At $u=\frac\pi2$: $-\cos\left(\frac\pi2\right)=0$.

Step 8: Calculate the final result

\[\int_{\frac{1}{\pi}}^{\frac{2}{\pi}} \frac{1}{x^2} \sin \frac{1}{x} \, dx = 1 - 0 = 1\]

Verification using the antiderivative:

The antiderivative of $\frac{1}{x^2}\sin\frac{1}{x}$ is $F(x) = \cos\left(\frac{1}{x}\right)$. Using the chain rule: $F'(x) = \frac{1}{x^2}\sin\left(\frac{1}{x}\right)$. Applying the Newton-Leibniz formula:

\[\int_{\frac{1}{\pi}}^{\frac{2}{\pi}} \frac{1}{x^2} \sin \frac{1}{x} \, dx = \left[\cos\left(\frac{1}{x}\right)\right]_{\frac{1}{\pi}}^{\frac{2}{\pi}} = \cos\left(\frac{\pi}{2}\right) - \cos(\pi) = 0-(-1)=1\]

This confirms our result!

Geometric interpretation:

The integral represents the area under the curve $y = \frac{1}{x^2}\sin\frac{1}{x}$ from $x = \frac{1}{\pi}$ to $x = \frac{2}{\pi}$. At $x=\frac1\pi$: $y=\pi^2\sin(\pi)=0$; at $x=\frac2\pi$: $y=\frac{\pi^2}{4}\sin\left(\frac\pi2\right)=\frac{\pi^2}{4}\approx2.467$. The substitution $u=\frac1x$ transformed this into the simpler integral $\int_{\pi/2}^\pi \sin u\,du$, representing a quarter sine wave's area. The result being exactly 1 is elegant.

Pattern recognition:

This follows a common pattern: when you see $\frac{1}{x^2}f\left(\frac{1}{x}\right)$, the substitution $u=\frac1x$ often simplifies the problem dramatically because $\frac{1}{x^2}\,dx=-du$.

Final answer:

\[\boxed{\int_{\frac{1}{\pi}}^{\frac{2}{\pi}} \frac{1}{x^2} \sin \frac{1}{x} \, dx = 1}\]