What is the integral of (3y^2-2sin(y)) ?

The integral of (3y^2-2sin(y)) is y^3+2cos(y)+C

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integral of (3y^2-2sin(y))

\int (3 y^{2} - 2 sin (y)) d y

y^{3} + 2 cos (y) + C

Solution steps

\int 3 y^{2} - 2 sin (y) d y

Apply the Sum Rule:

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

= \int 3 y^{2} d y - \int 2 sin (y) d y

\int 3 y^{2} d y = y^{3}

\int 2 sin (y) d y = - 2 cos (y)

= y^{3} - (- 2 cos (y))

Simplify

= y^{3} + 2 cos (y)

Add a constant to the solution

= y^{3} + 2 cos (y) + C

(x - 2)^{2} (x - 3)^{3}

\int_{0}^{\infty} \frac{x}{x ^{3}} d x

(5 x^{6} + 2 x^{3})^{4}

f (x) = x e^{x} csc (x)

\int 9 e^{\frac{x}{4}} d x

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