What is the integral of 4/(x^5)cos(1/(x^2)) ?

The integral of 4/(x^5)cos(1/(x^2)) is -2((sin(1/(x^2)))/(x^2)+cos(1/(x^2)))+C

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integral of 4/(x^5)cos(1/(x^2))

\int \frac{4}{x ^{5}} cos (\frac{1}{x ^{2}}) d x

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

Solution steps

\int \frac{4}{x ^{5}} cos (\frac{1}{x ^{2}}) 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 ^{5}} cos (\frac{1}{x ^{2}}) d x

Apply u-substitution

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

= 4 \cdot \int - \frac{u cos ( u )}{2} d u

Take the constant out:

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

= 4 (- \frac{1}{2} \cdot \int u cos (u) d u)

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= 4 (- \frac{1}{2} (u sin (u) - \int sin (u) d u))

Use the common integral:

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

= 4 (- \frac{1}{2} (u sin (u) - (- cos (u))))

Substitute back

u = \frac{1}{x ^{2}}

= 4 (- \frac{1}{2} (\frac{1}{x ^{2}} sin (\frac{1}{x ^{2}}) - (- cos (\frac{1}{x ^{2}}))))

Simplify

4 (- \frac{1}{2} (\frac{1}{x ^{2}} sin (\frac{1}{x ^{2}}) - (- cos (\frac{1}{x ^{2}})))) : - 2 (\frac{sin ( \frac{1}{x ^{2}} )}{x ^{2}} + cos (\frac{1}{x ^{2}}))

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

Add a constant to the solution

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

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What is the integral of 4/(x^5)cos(1/(x^2)) ?
The integral of 4/(x^5)cos(1/(x^2)) is -2((sin(1/(x^2)))/(x^2)+cos(1/(x^2)))+C