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

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

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

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

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \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}{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

= \frac{1}{2} (\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

= \frac{1}{2} (\frac{π}{2} + 0)

Simplify

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

= \frac{π}{4}

sin^{4} (\frac{x}{4}) + cos^{4} (\frac{x}{4})

\frac{\partial}{\partial x} (x y + y + 2 x + 2)

f (θ) = θ cos (θ) sin (θ)

\frac{\partial}{\partial y} (y^{2} - 4 y + x y)

\int (4 x^{2} - 9 x)^{5} (8 x - 9) d x

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