Evaluate the integral: I = integral sin 1/x · 1/(x²) dx

Evaluate the integral: $$ I = \int \sin \frac{1}{x} \cdot \frac{1}{x^2} \, dx. $$

Explanation of Method

Since the derivative of $\tfrac1x$ is $-\tfrac{1}{x^2}$, substituting $u=\tfrac1x$ converts the integrand into $-\sin u\,du$, a standard trigonometric antiderivative.

Substitute $u=\tfrac1x$.
Convert $\tfrac{1}{x^2}\,dx$ into $-du$.
Integrate using the standard sine antiderivative.
Substitute back and add the constant of integration.

Step 1 · Substitution

We use the substitution:

$$ u = \frac{1}{x} \quad \Rightarrow \quad du = -\frac{dx}{x^2}. $$

Step 2 · Rewrite and integrate

Rewriting the integral in terms of $u$:

$$ I = \int \sin u\cdot (-du). $$

Since the integral of $\sin u$ is $-\cos u$, we get:

$$ I = \cos u + C. $$

Step 3 · Substitute back

Substituting back $u = \dfrac1x$:

$$ I = \cos\frac{1}{x} + C. $$

Summary Table

Quantity	Value
Substitution	$u = 1/x$
$dx/x^2$ in terms of $du$	$-du$
Antiderivative	$\cos u$

Additional Notes

The domain of $I$ requires $x \neq 0$, since $1/x$ and $1/x^2$ are undefined there.
The double negative — from $du=-dx/x^2$ and $\int \sin u\,du=-\cos u$ — combines to give a positive $\cos u$; it's worth double-checking both signs carefully in problems like this.
The result can be checked by differentiating $\cos(1/x)$ with the chain rule and confirming it returns $\sin(1/x)\cdot \dfrac{1}{x^2}$.

Result

The solution to the integral is:

$$ I = \cos\frac{1}{x} + C. $$