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

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

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

\int_{0}^{\frac{π}{2}} cos (4 x) d x

0

Solution steps

\int_{0}^{\frac{π}{2}} cos (4 x) d x

Apply u-substitution

: \int_{0}^{2 π} cos (u) \frac{1}{4} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

= \frac{1}{4} [sin (u)]_{0}^{2 π}

Compute the boundaries:

0

= \frac{1}{4} \cdot 0

Simplify

= 0

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

\frac{d}{d x} (\frac{- x ^{2} + 9}{( x ^{2} + 9 ) ^{2}})

x \to 3 lim (- x^{2} - 4 x + 8)

\frac{d}{d x} (3 x^{2} + 6 x + 6)

\frac{d y}{d x} y = \frac{4 x}{3 x - 2 x ^{2}}

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