What is the integral of sin(8x+2) ?

The integral of sin(8x+2) is -1/8 cos(8x+2)+C

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integral of sin(8x+2)

\int sin (8 x + 2) d x

- \frac{1}{8} cos (8 x + 2) + C

Solution steps

\int sin (8 x + 2) d x

Apply u-substitution

: \int sin (u) \frac{1}{8} d u

= \int sin (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{8} (- cos (u))

Substitute back

u = 8 x + 2

= \frac{1}{8} (- cos (8 x + 2))

Simplify

= - \frac{1}{8} cos (8 x + 2)

Add a constant to the solution

= - \frac{1}{8} cos (8 x + 2) + C

2 \cdot y^{^{''}} + y^{^{'}} - y = 0

y^{^{''}} + 2 y^{^{'}} = 2 t + 5 - e^{- 2 t}

\frac{d}{d x} (x^{2 - 2 x})

y = \frac{ln ( x )}{e ^{x}}

\int csc (6 x) d x

What is the integral of sin(8x+2) ?
The integral of sin(8x+2) is -1/8 cos(8x+2)+C