What is the integral from 0 to pi/2 of (sin(2x))/2 ?

The integral from 0 to pi/2 of (sin(2x))/2 is 1/2

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integral from 0 to pi/2 of (sin(2x))/2

\int_{0}^{\frac{π}{2}} \frac{sin ( 2 x )}{2} d x

\frac{1}{2}

Solution steps

\int_{0}^{\frac{π}{2}} \frac{sin ( 2 x )}{2} 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_{0}^{\frac{π}{2}} sin (2 x) d x

Apply u-substitution

: \int_{0}^{π} sin (u) \frac{1}{2} d u

= \frac{1}{2} \cdot \int_{0}^{π} 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_{0}^{π} sin (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{π}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{π} : \frac{1}{4} [- cos (u)]_{0}^{π}

= \frac{1}{4} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{4} \cdot 2

Simplify

= \frac{1}{2}

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What is the integral from 0 to pi/2 of (sin(2x))/2 ?
The integral from 0 to pi/2 of (sin(2x))/2 is 1/2