What is the integral of-8sin(sqrt(x)) ?

The integral of-8sin(sqrt(x)) is -16(-sqrt(x)cos(sqrt(x))+sin(sqrt(x)))+C

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integral of-8sin(sqrt(x))

\int - 8 sin (x) d x

- 16 (- x cos (x) + sin (x)) + C

Solution steps

\int - 8 sin (x) d x

Take the constant out:

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

= - 8 \cdot \int sin (x) d x

Apply u-substitution

: \int sin (u) \cdot 2 u d u

= - 8 \cdot \int sin (u) \cdot 2 u d u

Take the constant out:

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

= - 8 \cdot 2 \cdot \int sin (u) u d u

Apply Integration By Parts

: - u cos (u) - \int - cos (u) d u

= - 8 \cdot 2 (- u cos (u) - \int - cos (u) d u)

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

= - 8 \cdot 2 (- u cos (u) - (- sin (u)))

Substitute back

u = x

= - 8 \cdot 2 (- x cos (x) - (- sin (x)))

Simplify

= - 16 (- x cos (x) + sin (x))

Add a constant to the solution

= - 16 (- x cos (x) + sin (x)) + C

\int - x^{4} + \frac{( 9 x ^{2} )}{4} - 2 x + 3 d x

(2 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0, y (1) = 5

\frac{\partial}{\partial x} ((1 - y) + 10 (x - y^{2})^{2})

x y^{^{'}} + 3 y = x^{2} - 1

\frac{\partial}{\partial x} (\frac{x}{y} + \frac{y}{x})

What is the integral of-8sin(sqrt(x)) ?
The integral of-8sin(sqrt(x)) is -16(-sqrt(x)cos(sqrt(x))+sin(sqrt(x)))+C