What is the integral of 2sin(6x) ?

The integral of 2sin(6x) is -1/3 cos(6x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2sin(6x)

\int 2 sin (6 x) d x

- \frac{1}{3} cos (6 x) + C

Solution steps

\int 2 sin (6 x) d x

Take the constant out:

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

= 2 \cdot \int sin (6 x) d x

Apply u-substitution

: \int sin (u) \frac{1}{6} d u

= 2 \cdot \int sin (u) \frac{1}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{6} (- cos (u))

Substitute back

u = 6 x

= 2 \cdot \frac{1}{6} (- cos (6 x))

Simplify

2 \cdot \frac{1}{6} (- cos (6 x)) : - \frac{1}{3} cos (6 x)

= - \frac{1}{3} cos (6 x)

Add a constant to the solution

= - \frac{1}{3} cos (6 x) + C

\frac{d}{d x} (25 x)

\int \frac{x ^{3} + 1}{x} d x

y^{^{'}} = 2 y^{2}

y (1 + x^{2}) y^{^{'}} - x (1 + y^{2}) = 0, y (0) = 3

\int \frac{3 ln ^{2} ( x ) + 36}{x} d x