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

The integral from 0 to pi/4 of (cos(2x)) is 1/2

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{\frac{π}{4}} cos (2 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{2} d u

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

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}} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{2} [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{1}{2} \cdot 1

Simplify

= \frac{1}{2}

y^{^{''}} + 2 y^{^{'}} + 5 y = - 2 t + 4 t^{2}

\frac{d}{d x} (\frac{3}{x ^{3}} - \frac{1}{x ^{4}})

\frac{d}{d x} (\frac{( x ^{2} - 3 x )}{( 2 x + 1 )})

3 y^{^{''}} + y = 0

\int_{0}^{1} 2 x \cdot 1 + 4 x^{2} d x

What is the integral from 0 to pi/4 of (cos(2x)) ?
The integral from 0 to pi/4 of (cos(2x)) is 1/2