Help with antiderivatives.?

Help with antiderivatives.?
What is the antiderivative of 2sin(x) - 8x ?
Thank you!
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answers:
Como say: ∫ 2 sin x - 8x dx = - 2 cos x - 4x² + C
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Vaman say: Anti derivative= integral. Check it your self
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Φ² = Φ+1 say: d(-2cosx)/dx = 2sinx
d(-4x²)/dx = -8x
d(C)/dx = 0
∫(2sinx - 8x)dx = -2cosx - 4x² + C
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Michael say: Well,

∫ (2sin x - 8x) dx = -2cos x - 4x² + C

hope it' ll help !!
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bramhananda r say: anti derivative of 2sin(x) - 8x
is integral of 2sin x -8x

integral fo sinx is - cos x dx

and of 8x is 8x^2/2 = 4x^2

so the anti derivtive of given function is

-2cosx -4x^2 +C
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Ian H say: -2cos(x) - 4x^2 + constant
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germano say: Hello,

let's evaluate the indefinite integral

∫ (2sinx - 8x) dx =

let's split into two integrals, factoring constants out:

2 ∫ sinx dx - 8 ∫ x dx =

2(- cosx) - 8 ∫ x dx =

(applying the integration rule ∫ xⁿ dx = [1/(n+1)] xⁿ⁺¹ + C)

- 2cosx - 8 [1/(1+1)] x¹ ⁺ ¹ + C =

- 2cosx - 8(1/2)x² + C =

- 2cosx - 4x² + C


the antiderivative is:

- 2cosx - 4x² + C



I hope it helps
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