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

1 Answer
Jan 31, 2015

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.