What is the integral of x^2sin(x^3+4) ?

The integral of x^2sin(x^3+4) is -1/3 cos(x^3+4)+C

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integral of x^2sin(x^3+4)

\int x^{2} sin (x^{3} + 4) d x

- \frac{1}{3} cos (x^{3} + 4) + C

Solution steps

\int x^{2} sin (x^{3} + 4) d x

Apply u-substitution

: \int \frac{sin ( u )}{3} d u

= \int \frac{sin ( u )}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{3} (- cos (u))

Substitute back

u = x^{3} + 4

= \frac{1}{3} (- cos (x^{3} + 4))

Simplify

= - \frac{1}{3} cos (x^{3} + 4)

Add a constant to the solution

= - \frac{1}{3} cos (x^{3} + 4) + C

x \to c lim (x)

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

y = 4 - x^{2}, y = - 3 x

\int_{1}^{2} π (\frac{1}{x})^{2} d x

\int_{5}^{11} \frac{16}{x - 1} d x

What is the integral of x^2sin(x^3+4) ?
The integral of x^2sin(x^3+4) is -1/3 cos(x^3+4)+C