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

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

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integral of sin(1/3 x)

\int sin (\frac{1}{3} x) d x

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

Solution steps

\int sin (\frac{1}{3} x) d x

Apply u-substitution

: \int sin (u) \cdot 3 d u

= \int sin (u) \cdot 3 d u

Take the constant out:

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

= 3 \cdot \int sin (u) d u

Use the common integral:

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

= 3 (- cos (u))

Substitute back

u = \frac{1}{3} x

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (2 sin (3 x))

\frac{d}{d x} (cos^{3} (3 x))

f (x) = 3 x + 7

y = 3^{x - 1}

f (x) = \frac{x}{x ^{2} + 9}

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