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 pi/4

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

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

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} 4 cos^{2} (x) sin^{2} (x) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (x) sin^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 x )}{8} d x

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d x - \int_{0}^{\frac{π}{2}} cos (4 x) d x)

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 x) d x = 0

= 4 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

4 \cdot \frac{1}{8} (\frac{π}{2} - 0) : \frac{π}{4}

= \frac{π}{4}

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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 pi/4