What is the integral of sin(6x-10) ?

The integral of sin(6x-10) is -1/6 cos(6x-10)+C

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integral of sin(6x-10)

\int sin (6 x - 10) d x

- \frac{1}{6} cos (6 x - 10) + C

Solution steps

\int sin (6 x - 10) d x

Apply u-substitution

: \int sin (u) \frac{1}{6} d u

= \int sin (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{6} (- cos (u))

Substitute back

u = 6 x - 10

= \frac{1}{6} (- cos (6 x - 10))

Simplify

= - \frac{1}{6} cos (6 x - 10)

Add a constant to the solution

= - \frac{1}{6} cos (6 x - 10) + C

f (x) = x^{2} + 2, (- 4, 18)

\frac{d}{d x} (sec^{2} (x) + csc^{2} (x))

\frac{d}{d x} (g (h, x) (h (x)))

\int \frac{cos ( x )}{7 + sin ^{2} ( x )} d x

f (x) = (x + 2)^{\frac{2}{3}} (2 x^{2} - 5)^{3}

What is the integral of sin(6x-10) ?
The integral of sin(6x-10) is -1/6 cos(6x-10)+C