What is the integral of sin(4x-2) ?

The integral of sin(4x-2) is -1/4 cos(4x-2)+C

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integral of sin(4x-2)

\int sin (4 x - 2) d x

- \frac{1}{4} cos (4 x - 2) + C

Solution steps

\int sin (4 x - 2) d x

Apply u-substitution

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

= \int sin (u) \frac{1}{4} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} (- cos (u))

Substitute back

u = 4 x - 2

= \frac{1}{4} (- cos (4 x - 2))

Simplify

= - \frac{1}{4} cos (4 x - 2)

Add a constant to the solution

= - \frac{1}{4} cos (4 x - 2) + C

h (x) = x cos (x) + 3 π

x \to 4 lim (cos (\frac{π x}{3}))

\int \frac{4 e ^{x}}{e ^{2 x} + 6 e ^{x} + 9} d x

y = 3.6 x + 500

\frac{d}{d x} (2 x^{3} - 15 x^{2} - 36 x)

What is the integral of sin(4x-2) ?
The integral of sin(4x-2) is -1/4 cos(4x-2)+C