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

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

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integral of 6sin(3x)

\int 6 sin (3 x) d x

- 2 cos (3 x) + C

Solution steps

\int 6 sin (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 sin (3 x) 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

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

Simplify

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

= - 2 cos (3 x)

Add a constant to the solution

= - 2 cos (3 x) + C

\frac{x ^{2} - 3}{( x - 1 ) ^{2} + ( x - 1 ) ^{2}}

\int_{1}^{4} \frac{1}{9 + ( x - 1 ) ^{2}} d x

\frac{d}{d x} (24 x - 4 x)

\frac{8 x}{x}