What is the integral of 4/(x^2)cos(4/x) ?

The integral of 4/(x^2)cos(4/x) is -sin(4/x)+C

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integral of 4/(x^2)cos(4/x)

\int \frac{4}{x ^{2}} cos (\frac{4}{x}) d x

- sin (\frac{4}{x}) + C

Solution steps

\int \frac{4}{x ^{2}} cos (\frac{4}{x}) d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2}} cos (\frac{4}{x}) d x

Apply u-substitution

= 4 \cdot \int - \frac{cos ( u )}{4} d u

Take the constant out:

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

= 4 (- \frac{1}{4} \cdot \int cos (u) d u)

Use the common integral:

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

= 4 (- \frac{1}{4} sin (u))

Substitute back

u = \frac{4}{x}

= 4 (- \frac{1}{4} sin (\frac{4}{x}))

Simplify

4 (- \frac{1}{4} sin (\frac{4}{x})) : - sin (\frac{4}{x})

= - sin (\frac{4}{x})

Add a constant to the solution

= - sin (\frac{4}{x}) + C

\int_{0}^{π} sin (x) sin (n x) d x

\int (4 x - 3)^{6} d x

t an g e n t f (x) = \frac{6}{x}, at x = 1

x \to 4 - lim (\frac{x - 4}{x - 1})

d er i v a t i v e y = \frac{1}{x}

What is the integral of 4/(x^2)cos(4/x) ?
The integral of 4/(x^2)cos(4/x) is -sin(4/x)+C