What is the integral of 3cos(4x-1) ?

The integral of 3cos(4x-1) is 3/4 sin(4x-1)+C

Popular >

integral of 3cos(4x-1)

\int 3 cos (4 x - 1) d x

\frac{3}{4} sin (4 x - 1) + C

Solution steps

\int 3 cos (4 x - 1) d x

Take the constant out:

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

= 3 \cdot \int cos (4 x - 1) d x

Apply u-substitution

: \int cos (u) \frac{1}{4} d u

= 3 \cdot \int cos (u) \frac{1}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int cos (u) d u

Use the common integral:

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

= 3 \cdot \frac{1}{4} sin (u)

Substitute back

u = 4 x - 1

= 3 \cdot \frac{1}{4} sin (4 x - 1)

Simplify

3 \cdot \frac{1}{4} sin (4 x - 1) : \frac{3}{4} sin (4 x - 1)

= \frac{3}{4} sin (4 x - 1)

Add a constant to the solution

= \frac{3}{4} sin (4 x - 1) + C

\int - \frac{3 ( x - 6 )}{x ^{2} - 17 x + 72} d x

\frac{3}{s ( s ^{2} + 6 s )}

x \to 5 lim (- x^{2} + 2 x - 5)

\int_{0}^{10} 400 - 20 t d t

x \to 0 + lim (lo g_{\frac{1}{2}} (x))

What is the integral of 3cos(4x-1) ?
The integral of 3cos(4x-1) is 3/4 sin(4x-1)+C