What is the integral of sin(1/(2x)) ?

The integral of sin(1/(2x)) is xsin(1/(2x))-1/2 \Ci(1/(2x))+C

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integral of sin(1/(2x))

\int sin (\frac{1}{2 x}) d x

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

Solution steps

\int sin (\frac{1}{2 x}) d x

Apply u-substitution

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

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

Take the constant out:

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

= - \frac{1}{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

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

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

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

Substitute back

u = \frac{1}{2 x}

= - \frac{1}{2} (- \frac{sin ( \frac{1}{2 x} )}{\frac{1}{2 x}} - (- Ci (\frac{1}{2 x})))

Simplify

- \frac{1}{2} (- \frac{sin ( \frac{1}{2 x} )}{\frac{1}{2 x}} - (- Ci (\frac{1}{2 x}))) : x sin (\frac{1}{2 x}) - \frac{1}{2} Ci (\frac{1}{2 x})

= x sin (\frac{1}{2 x}) - \frac{1}{2} Ci (\frac{1}{2 x})

Add a constant to the solution

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

y = 16 x

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What is the integral of sin(1/(2x)) ?
The integral of sin(1/(2x)) is xsin(1/(2x))-1/2 \Ci(1/(2x))+C