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

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

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

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

\frac{π + 2}{16}

Decimal

0.32134 \dots

Solution steps

\int_{0}^{\frac{π}{8}} cos^{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}

x^{5} cos (x^{2})

\frac{d}{d x} ((3 x)^{5 x})

\int a x + b d x

\frac{d}{d x} (in (x in x))

\frac{d}{d x} (11 - x + 4 ln (2 x - 1))

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