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

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

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

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

Simplify

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

Rewrite using trig identities

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

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 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}

n = 1 \sum \infty 4^{\frac{1}{2 ^{n}}}

y = (2^{x})^{8}

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

4 y^{^{''}} + 4 y^{^{'}} + 8 y = 0

x \to 0 lim (cos (9 x))

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