What is the integral of (sin(1/(x^3)))/(x^4) ?

The integral of (sin(1/(x^3)))/(x^4) is 1/3 cos(1/(x^3))+C

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integral of (sin(1/(x^3)))/(x^4)

\int \frac{sin ( \frac{1}{x ^{3}} )}{x ^{4}} d x

\frac{1}{3} cos (\frac{1}{x ^{3}}) + C

Solution steps

\int \frac{sin ( \frac{1}{x ^{3}} )}{x ^{4}} d x

Apply u-substitution

: \int \frac{sin ( \frac{1}{u} )}{3 u ^{2}} d u

= \int \frac{sin ( \frac{1}{u} )}{3 u ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{sin ( \frac{1}{u} )}{u ^{2}} d u

Apply u-substitution

: \int - sin (v) d v

= \frac{1}{3} \cdot \int - sin (v) d v

Take the constant out:

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

= \frac{1}{3} (- \int sin (v) d v)

Use the common integral:

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

= \frac{1}{3} (- (- cos (v)))

Substitute back

= \frac{1}{3} (- (- cos (\frac{1}{x ^{3}})))

Simplify

= \frac{1}{3} cos (\frac{1}{x ^{3}})

Add a constant to the solution

= \frac{1}{3} cos (\frac{1}{x ^{3}}) + C

\frac{\partial}{\partial x} (\frac{( 4 x - 4 y )}{4 x + 4 y})

\frac{\partial}{\partial x} (tan (8 + 2 x^{2} y^{4} z^{2}))

\frac{\partial}{\partial x} (324 x^{2} - 324 x y + 81 y^{2})

4 cos (3 x), 4 sin (6 x), 0, \frac{π}{6}

8 arcsin (x^{4})

What is the integral of (sin(1/(x^3)))/(x^4) ?
The integral of (sin(1/(x^3)))/(x^4) is 1/3 cos(1/(x^3))+C