What is the integral of 6sin(3x+1) ?

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

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integral of 6sin(3x+1)

\int 6 sin (3 x + 1) d x

- 2 cos (3 x + 1) + C

Solution steps

\int 6 sin (3 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 (3 x + 1) d x

Apply u-substitution

: \int sin (u) \frac{1}{3} d u

= 6 \cdot \int sin (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 sin (u) d u

Use the common integral:

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

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

Substitute back

u = 3 x + 1

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

Simplify

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

= - 2 cos (3 x + 1)

Add a constant to the solution

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

x \to \frac{1}{6} lim (∣ - 6 x + 1 ∣)

\frac{d}{d x} (x^{5} - 5 x^{4})

\frac{d}{d x} (\frac{x}{1 - 4 x ^{2}})

\frac{\partial}{\partial y} (y^{2} ln (x))

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

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