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

The integral of 3cos(1/3 x) is 9sin(1/3 x)+C

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integral of 3cos(1/3 x)

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

9 sin (\frac{1}{3} x) + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int cos (\frac{1}{3} x) d x

Apply u-substitution

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

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

Use the common integral:

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

= 3 \cdot 3 sin (u)

Substitute back

u = \frac{1}{3} x

= 3 \cdot 3 sin (\frac{1}{3} x)

Simplify

= 9 sin (\frac{1}{3} x)

Add a constant to the solution

= 9 sin (\frac{1}{3} x) + C

\int x^{8} - 4 x^{- 5} + 6 d x

\frac{\partial}{\partial y} (y z^{2})

x \to 0 lim (x^{2} cos (\frac{1}{x}))

f (x) sin (x)

\frac{d y}{d x} = \frac{( 3 x + 4 y )}{( 3 x + 4 y + 4 )}

What is the integral of 3cos(1/3 x) ?
The integral of 3cos(1/3 x) is 9sin(1/3 x)+C