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

The integral from 0 to pi/8 of sin^2(2x) is (pi-2)/(16)

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

\int_{0}^{\frac{π}{8}} sin^{2} (2 x) d x

\frac{π - 2}{16}

Decimal Notation

0.07134 \dots

Solution steps

\int_{0}^{\frac{π}{8}} sin^{2} (2 x) d x

Rewrite using trig identities

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

\int_{0}^{\frac{π}{8}} 1 d x = \frac{π}{8}

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

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

Simplify

\frac{1}{2} (\frac{π}{8} - \frac{1}{4}) : \frac{π - 2}{16}

= \frac{π - 2}{16}

sin (x) cos (x)

\int x^{3} (x^{4} + 3)^{2} d x

y = x^{5} - 4 x^{3} + 2 x - 3

\frac{8 x}{1 - x}

- 2 t + 8

What is the integral from 0 to pi/8 of sin^2(2x) ?
The integral from 0 to pi/8 of sin^2(2x) is (pi-2)/(16)