What is the integral from 0 to pi/2 of cos(3x) ?

The integral from 0 to pi/2 of cos(3x) is -1/3

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integral from 0 to pi/2 of cos(3x)

\int_{0}^{\frac{π}{2}} cos (3 x) d x

- \frac{1}{3}

Decimal Notation

- 0.33333 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos (3 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

- 1

= \frac{1}{3} (- 1)

Simplify

= - \frac{1}{3}

\int (2 e^{x} + 8 x^{3} + 6 x) d x

\int \frac{4}{15 sin ( x ) - 8 cos ( x )} d x

y = 4 - x^{2}

\frac{d y}{d x} = \frac{x ^{2} - 1}{y ^{2} + 1}, y (- 1) = 1

n \to \infty lim (\frac{n}{- n})

What is the integral from 0 to pi/2 of cos(3x) ?
The integral from 0 to pi/2 of cos(3x) is -1/3