What is the integral of 6/(x^2)cos(6/x) ?

The integral of 6/(x^2)cos(6/x) is -sin(6/x)+C

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integral of 6/(x^2)cos(6/x)

\int \frac{6}{x ^{2}} cos (\frac{6}{x}) d x

- sin (\frac{6}{x}) + C

Solution steps

\int \frac{6}{x ^{2}} cos (\frac{6}{x}) d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ^{2}} cos (\frac{6}{x}) d x

Apply u-substitution

: \int - \frac{cos ( u )}{6} d u

= 6 \cdot \int - \frac{cos ( u )}{6} d u

Take the constant out:

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

= 6 (- \frac{1}{6} \cdot \int cos (u) d u)

Use the common integral:

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

= 6 (- \frac{1}{6} sin (u))

Substitute back

u = \frac{6}{x}

= 6 (- \frac{1}{6} sin (\frac{6}{x}))

Simplify

6 (- \frac{1}{6} sin (\frac{6}{x})) : - sin (\frac{6}{x})

= - sin (\frac{6}{x})

Add a constant to the solution

= - sin (\frac{6}{x}) + C

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What is the integral of 6/(x^2)cos(6/x) ?
The integral of 6/(x^2)cos(6/x) is -sin(6/x)+C