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

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Answer 1 · 2015-01-31T18:58:27

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#.