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

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

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

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

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

Solution steps

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

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 2})

\frac{\partial}{\partial x} ((6 - x - y)^{2})

\frac{d y}{d t} = - 2 (y - 3)

\frac{\partial}{\partial x} (17 x y e^{- x^{2} - y^{2}})

\frac{\partial}{\partial x} (c (1 + m \cdot x) - c (1 + m)^{x})

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