What is the integral of (cos(1/x))/(x^2) ?

The integral of (cos(1/x))/(x^2) is -sin(1/x)+C

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integral of (cos(1/x))/(x^2)

\int \frac{cos ( \frac{1}{x} )}{x ^{2}} d x

- sin (\frac{1}{x}) + C

Solution steps

\int \frac{cos ( \frac{1}{x} )}{x ^{2}} d x

Apply u-substitution

: \int - cos (u) d u

= \int - cos (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= - sin (u)

Substitute back

u = \frac{1}{x}

= - sin (\frac{1}{x})

Add a constant to the solution

= - sin (\frac{1}{x}) + C

(lo g_{10} (n))^{2}

x y^{^{'}} - y = \frac{3}{2} x ln (x)

\int \frac{5 x - 1}{x ^{2} - 3 x - 40} d x

f (x) = \frac{x}{8 - x}, at x = 3

\frac{\partial}{\partial x} (2 x e^{\frac{x}{2} + y} + 2)

What is the integral of (cos(1/x))/(x^2) ?
The integral of (cos(1/x))/(x^2) is -sin(1/x)+C