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

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

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

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

\frac{π + 2}{8}

Decimal

0.64269 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \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}{4}} 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}{4}} 1 d x + \int_{0}^{\frac{π}{4}} cos (2 x) d x)

\int_{0}^{\frac{π}{4}} 1 d x = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 x) d x = \frac{1}{2}

= \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Simplify

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

= \frac{π + 2}{8}

\frac{\partial}{\partial y} (4 e^{y})

4 y^{^{''}} + 7 y^{^{'}} - 2 y = 0

\int \frac{( x + 1 )}{x ^{2}} d x

\frac{d}{d y} (ln (y) - \frac{y ^{2}}{8})

\int cos (4 x) d x

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