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_{- 1}^{1} e^{x + 1} d x

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

\frac{\partial}{\partial y} (x^{4} - 3 x^{2} y^{2} + y^{4})

\int \frac{cos ( x )}{sin ( x ) + 1} d x

\frac{d}{d x} (\frac{sin ( x )}{2})

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