What is the integral from 0 to pi/4 of cos(6x) ?

The integral from 0 to pi/4 of cos(6x) is -1/6

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integral from 0 to pi/4 of cos(6x)

\int_{0}^{\frac{π}{4}} cos (6 x) d x

- \frac{1}{6}

Decimal Notation

- 0.16666 \dots

Solution steps

\int_{0}^{\frac{π}{4}} cos (6 x) d x

Apply u-substitution

: \int_{0}^{\frac{3 π}{2}} cos (u) \frac{1}{6} d u

= \int_{0}^{\frac{3 π}{2}} cos (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{\frac{3 π}{2}} cos (u) d u

Use the common integral:

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

= \frac{1}{6} [sin (u)]_{0}^{\frac{3 π}{2}}

Compute the boundaries:

- 1

= \frac{1}{6} (- 1)

Simplify

= - \frac{1}{6}

y^{^{'}} (e^{x} - 1) ln (y) = y

- \frac{16}{( 8 x + 9 ) ^{3}}

\frac{d y}{d x} = \frac{y ^{2} + y}{x ^{2} + x}

\int_{0}^{1} x e^{- 2 x} d x

y = 4 x^{2} - 1, y = 0, x = \frac{- 1}{2}, x = 1

What is the integral from 0 to pi/4 of cos(6x) ?
The integral from 0 to pi/4 of cos(6x) is -1/6