What is the integral from 0 to 1 of 6cos(3x) ?

The integral from 0 to 1 of 6cos(3x) is 2sin(3)

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integral from 0 to 1 of 6cos(3x)

\int_{0}^{1} 6 cos (3 x) d x

2 sin (3)

Decimal Notation

0.28224 \dots

Solution steps

\int_{0}^{1} 6 cos (3 x) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} cos (3 x) d x

Apply u-substitution

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

= 6 \cdot \int_{0}^{3} 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

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

Use the common integral:

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

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

Simplify

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

= 2 [sin (u)]_{0}^{3}

Compute the boundaries:

sin (3)

= 2 sin (3)

x \to 1 - lim (\frac{2}{ln ( x )})

(0, 9), (7, 4)

\frac{d}{d x} (7 x^{2} tan (x))

6 t^{3} e^{- 2 t}

\int \frac{1}{( 1 + 4 x ^{2} ) ( arctan ( 2 x ) ) ^{3}} d x

What is the integral from 0 to 1 of 6cos(3x) ?
The integral from 0 to 1 of 6cos(3x) is 2sin(3)