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

Jan 31, 2015

The answer is: $= {x}^{3} - \frac{1}{4} \cos 4 x + \frac{1}{\cos} x + c$

Because:

$\int \left(3 {x}^{2} + \sin 4 x + \tan x \sec x\right) \mathrm{dx} =$

$= \int 3 {x}^{2} \mathrm{dx} + \frac{1}{4} \int 4 \sin 4 x \mathrm{dx} + \int \sin \frac{x}{\cos} x \cdot \frac{1}{\cos} x \mathrm{dx} =$

$= 3 {x}^{3} / 3 + \frac{1}{4} \left(- \cos 4 x\right) + \int \sin x \cdot {\cos}^{-} 2 x \mathrm{dx} =$

$= {x}^{3} - \frac{1}{4} \cos 4 x - \frac{{\cos}^{- 2 + 1} x}{- 2 + 1} + c =$

$= {x}^{3} - \frac{1}{4} \cos 4 x + {\cos}^{-} 1 x + c =$

$= {x}^{3} - \frac{1}{4} \cos 4 x + \frac{1}{\cos} x + c$.

To do these integrals i have used these rules:

$\int {x}^{n} \mathrm{dx} = {n}^{n + 1} / \left(n + 1\right) + c$,

$\int \sin f \left(x\right) f ' \left(x\right) \mathrm{dx} = - \cos f \left(x\right) + c$,

$\int {\left[f \left(x\right)\right]}^{n} f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c$.