What is the integral from 0 to pi/4 of (cos(2x))^2 ?

The integral from 0 to pi/4 of (cos(2x))^2 is pi/8

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integral from 0 to pi/4 of (cos(2x))^2

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

\frac{π}{8}

Decimal

0.39269 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \frac{1}{2} (1 + cos (4 x)) 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}{4}} 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

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

\int_{0}^{\frac{π}{4}} 1 d x = \frac{π}{4}

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

= \frac{1}{2} (\frac{π}{4} + 0)

Simplify

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

= \frac{π}{8}

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What is the integral from 0 to pi/4 of (cos(2x))^2 ?
The integral from 0 to pi/4 of (cos(2x))^2 is pi/8