What is the integral of 8x^3-cos(4x) ?

The integral of 8x^3-cos(4x) is 2x^4-1/4 sin(4x)+C

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integral of 8x^3-cos(4x)

\int 8 x^{3} - cos (4 x) d x

2 x^{4} - \frac{1}{4} sin (4 x) + C

Solution steps

\int 8 x^{3} - cos (4 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 x^{3} d x - \int cos (4 x) d x

\int 8 x^{3} d x = 2 x^{4}

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= 2 x^{4} - \frac{1}{4} sin (4 x)

Add a constant to the solution

= 2 x^{4} - \frac{1}{4} sin (4 x) + C

e

x^{2} \frac{d y}{d x} = x y + y^{2}

f (x) = \frac{3}{x}, (- 8, - \frac{3}{8})

\frac{d}{d h} (\frac{e ^{ah} - 1}{h})

x \to 4 lim (x ∣ x - 4 ∣)

What is the integral of 8x^3-cos(4x) ?
The integral of 8x^3-cos(4x) is 2x^4-1/4 sin(4x)+C