What is the integral of sin(3x+5) ?

The integral of sin(3x+5) is -1/3 cos(3x+5)+C

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integral of sin(3x+5)

\int sin (3 x + 5) d x

- \frac{1}{3} cos (3 x + 5) + C

Solution steps

\int sin (3 x + 5) d x

Apply u-substitution

: \int sin (u) \frac{1}{3} d u

= \int sin (u) \frac{1}{3} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} (- cos (u))

Substitute back

u = 3 x + 5

= \frac{1}{3} (- cos (3 x + 5))

Simplify

= - \frac{1}{3} cos (3 x + 5)

Add a constant to the solution

= - \frac{1}{3} cos (3 x + 5) + C

\frac{\partial}{\partial y} (x^{2} - 2 y)

4 e^{- 3 t} cos (2 t)

t \to 0 lim (\frac{9}{t} - \frac{9}{t ^{2} + t})

(y - x) y^{^{'}} = y - x + 3

\int \frac{1}{1 + w ^{2}} d w

What is the integral of sin(3x+5) ?
The integral of sin(3x+5) is -1/3 cos(3x+5)+C