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

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

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

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

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

Solution steps

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

Simplify

\frac{1}{x ^{3}} sin (\frac{1}{x}) : \frac{sin ( \frac{1}{x} )}{x ^{3}}

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

Apply u-substitution

: \int - u sin (u) d u

= \int - u sin (u) d u

Take the constant out:

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

= - \int u sin (u) d u

Apply Integration By Parts

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

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

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

= - (- u cos (u) - (- sin (u)))

Substitute back

u = \frac{1}{x}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 24 t e^{2 t}, y (0) = 5, y^{^{'}} (0) = 0

\frac{d}{d x} (x (4))

y = \frac{4 x}{x + 1}, (3, 3)

6 x^{2} + 4 x - 3

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