What is the integral of (4x^2-3cos(x)) ?

The integral of (4x^2-3cos(x)) is (4x^3)/3-3sin(x)+C

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integral of (4x^2-3cos(x))

\int (4 x^{2} - 3 cos (x)) d x

\frac{4 x ^{3}}{3} - 3 sin (x) + C

Solution steps

\int 4 x^{2} - 3 cos (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 x^{2} d x - \int 3 cos (x) d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 3 cos (x) d x = 3 sin (x)

= \frac{4 x ^{3}}{3} - 3 sin (x)

Add a constant to the solution

= \frac{4 x ^{3}}{3} - 3 sin (x) + C

\int sin (10 x) cos (5 x) d x

f (x) = e^{l n (s i n (x^{2}))} 2 x + 1

\int e^{4 x} cos (6 x) d x

\frac{d x}{d t} = 3 x t^{2}

x \to - 3 lim (- \frac{6}{x + 3})

What is the integral of (4x^2-3cos(x)) ?
The integral of (4x^2-3cos(x)) is (4x^3)/3-3sin(x)+C