What is the integral of 1/2 sin(2θ) ?

The integral of 1/2 sin(2θ) is -1/4 cos(2θ)+C

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integral of 1/2 sin(2θ)

\int \frac{1}{2} sin (2 θ) d θ

- \frac{1}{4} cos (2 θ) + C

Solution steps

\int \frac{1}{2} sin (2 θ) d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (2 θ) d θ

Apply u-substitution

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

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

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} (- cos (u))

Substitute back

u = 2 θ

= \frac{1}{2} \cdot \frac{1}{2} (- cos (2 θ))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- cos (2 θ)) : - \frac{1}{4} cos (2 θ)

= - \frac{1}{4} cos (2 θ)

Add a constant to the solution

= - \frac{1}{4} cos (2 θ) + C

\frac{\partial}{\partial x} (x^{2} + 2 y^{2} - x^{2} y - 1)

y^{^{''}} + 81 y = cos (4 t)

\frac{d}{d t} (\frac{1}{3} π r (t) (t)^{2} h)

\frac{\partial}{\partial x} (\frac{7 x - 5 y}{y ^{2} + x ^{2}})

x \to - π lim ((x + π) csc (x))

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