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

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

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

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

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

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

f (x) = ln (sin (x))

15 x \cdot ln (20 x)

\int_{0}^{49} \frac{1}{x} d x

x \to 1 lim ((x^{2} - 9)^{\frac{1}{3}})

\frac{d}{d x} (x e^{x^{5}})

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