What is the integral of 6sin(2x+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6sin(2x+1)

\int 6 sin (2 x + 1) d x

- 3 cos (2 x + 1) + C

Solution steps

\int 6 sin (2 x + 1) d x

Take the constant out:

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

= 6 \cdot \int sin (2 x + 1) d x

Apply u-substitution

: \int sin (u) \frac{1}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 x + 1

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

Simplify

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

= - 3 cos (2 x + 1)

Add a constant to the solution

= - 3 cos (2 x + 1) + C

\int \frac{21}{x 1 + 2 x} d x

\int \frac{16}{x ^{2} - 16} d x

\int \frac{4 x ^{2}}{e ^{x^{3}}} d x

\int 3 x sec^{2} (3 x) d x

\int x sec^{2} (x^{2}) tan^{4} (x^{2}) d x

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