What is the integral of (cos(1/(x^2)))/(x^3) ?

The integral of (cos(1/(x^2)))/(x^3) is -1/2 sin(1/(x^2))+C

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integral of (cos(1/(x^2)))/(x^3)

\int \frac{cos ( \frac{1}{x ^{2}} )}{x ^{3}} d x

- \frac{1}{2} sin (\frac{1}{x ^{2}}) + C

Solution steps

\int \frac{cos ( \frac{1}{x ^{2}} )}{x ^{3}} d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{cos ( \frac{1}{u} )}{u ^{2}} d u

Apply u-substitution

: \int - cos (v) d v

= \frac{1}{2} \cdot \int - cos (v) d v

Take the constant out:

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

= \frac{1}{2} (- \int cos (v) d v)

Use the common integral:

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

= \frac{1}{2} (- sin (v))

Substitute back

= \frac{1}{2} (- sin (\frac{1}{x ^{2}}))

Simplify

= - \frac{1}{2} sin (\frac{1}{x ^{2}})

Add a constant to the solution

= - \frac{1}{2} sin (\frac{1}{x ^{2}}) + C

\int_{0}^{9} f (x) d x

x (\frac{d y}{d x}) - (1 + x) y = x y^{2}

3 e^{7}

\frac{d}{d x} (\frac{1 - 5 x}{1 + 5 x})

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

What is the integral of (cos(1/(x^2)))/(x^3) ?
The integral of (cos(1/(x^2)))/(x^3) is -1/2 sin(1/(x^2))+C