What is the integral of (-8cos(x)-2x) ?

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

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integral of (-8cos(x)-2x)

\int (- 8 cos (x) - 2 x) d x

- 8 sin (x) - x^{2} + C

Solution steps

\int - 8 cos (x) - 2 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 8 cos (x) d x - \int 2 x d x

\int 8 cos (x) d x = 8 sin (x)

\int 2 x d x = x^{2}

= - 8 sin (x) - x^{2}

Add a constant to the solution

= - 8 sin (x) - x^{2} + C

\int \frac{1}{16 x ^{2} - 49} d x

\int sin (8 - x) d x

\int_{- 1}^{0} x^{2} d x

\int e^{c o s (8 x)} sin (8 x) d x

x \to 1 lim ((x - 1) (x + 1))

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