What is the integral of 1/2 sin((2pi)/5 t) ?

The integral of 1/2 sin((2pi)/5 t) is -5/(4pi)cos((2pit)/5)+C

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integral of 1/2 sin((2pi)/5 t)

\int \frac{1}{2} sin (\frac{2 π}{5} t) d t

- \frac{5}{4 π} cos (\frac{2 π t}{5}) + C

Solution steps

\int \frac{1}{2} sin (\frac{2 π}{5} t) d t

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (\frac{2 π}{5} t) d t

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1}{2} \cdot \int sin (\frac{2 π t}{5}) d t

Apply u-substitution

= \frac{1}{2} \cdot \int sin (\frac{u}{5}) \frac{1}{2 π} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot \int sin (\frac{u}{5}) d u

Apply u-substitution

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot \int sin (v) \cdot 5 d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot 5 \cdot \int sin (v) d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot 5 (- cos (v))

Substitute back

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot 5 (- cos (\frac{2 π t}{5}))

Simplify

\frac{1}{2} \cdot \frac{1}{2 π} \cdot 5 (- cos (\frac{2 π t}{5})) : - \frac{5}{4 π} cos (\frac{2 π t}{5})

= - \frac{5}{4 π} cos (\frac{2 π t}{5})

Add a constant to the solution

= - \frac{5}{4 π} cos (\frac{2 π t}{5}) + C

What is the integral of 1/2 sin((2pi)/5 t) ?
The integral of 1/2 sin((2pi)/5 t) is -5/(4pi)cos((2pit)/5)+C