What is the integral of sin(2/x) ?

The integral of sin(2/x) is xsin(2/x)-2\Ci(2/x)+C

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integral of sin(2/x)

\int sin (\frac{2}{x}) d x

x sin (\frac{2}{x}) - 2 Ci (\frac{2}{x}) + C

Solution steps

\int sin (\frac{2}{x}) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

= - 2 (- \frac{sin ( u )}{u} - (- Ci (u)))

Substitute back

u = \frac{2}{x}

= - 2 (- \frac{sin ( \frac{2}{x} )}{\frac{2}{x}} - (- Ci (\frac{2}{x})))

Simplify

- 2 (- \frac{sin ( \frac{2}{x} )}{\frac{2}{x}} - (- Ci (\frac{2}{x}))) : x sin (\frac{2}{x}) - 2 Ci (\frac{2}{x})

= x sin (\frac{2}{x}) - 2 Ci (\frac{2}{x})

Add a constant to the solution

= x sin (\frac{2}{x}) - 2 Ci (\frac{2}{x}) + C

\int_{0}^{3} (6 x - x^{2}) d x

\frac{\partial}{\partial x} (x y + 3 x)

(1 + 4 x^{2}) y^{^{'}} = 1

e^{- k x}

arccot (2 y)

What is the integral of sin(2/x) ?
The integral of sin(2/x) is xsin(2/x)-2\Ci(2/x)+C