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

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

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integral of 1/2 sin((2pix)/5)

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

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

Solution steps

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

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 x}{5}) d x

Apply u-substitution

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

= \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

: \int sin (v) \cdot 5 d v

= \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 x}{5}))

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d t} = 3 y - e^{(3 t)} + 6 t^{2} - 2

\frac{d}{d x} ((5 x + 8)^{2})

y^{^{''}} + 25 y = 5 csc^{2} (5 t)

\frac{d}{d x} (x^{3 s i n (x)})

\int e^{y t} \cdot sin (k t) d t

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