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How do I find the antiderivative of #f(x) =3x^2 + sin(4x)+tan x sec x#?

Answer 1

Cameron Alexander

The answer is: #=x^3-1/4cos4x+1/cosx+c#

Because:

#int (3x^2+sin4x+tanxsecx)dx=#

#=int3x^2dx+1/4int4sin4xdx+intsinx/cosx*1/cosxdx=#

#=3x^3/3+1/4(-cos4x)+intsinx*cos^-2xdx=#

#=x^3-1/4cos4x-(cos^(-2+1)x)/(-2+1)+c=#

#=x^3-1/4cos4x+cos^-1x+c=#

#=x^3-1/4cos4x+1/cosx+c#.

To do these integrals i have used these rules:

#intx^ndx=n^(n+1)/(n+1)+c#,

#intsinf(x)f'(x)dx=-cosf(x)+c#,

#int[f(x)]^nf'(x)dx=[f(x)]^(n+1)/(n+1)+c#.

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Answer 2

Adam Adkins

To find the antiderivative of f(x) = 3x^2 + sin(4x) + tan(x)sec(x), you can integrate each term separately. The antiderivative of 3x^2 with respect to x is x^3, the antiderivative of sin(4x) with respect to x is -(1/4)cos(4x), and the antiderivative of tan(x)sec(x) with respect to x is sec(x). So, the antiderivative of f(x) is:

∫f(x) dx = x^3 - (1/4)cos(4x) + sec(x) + C

Where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.