What is the integral of (x+2)sin(x^2+4x-6) ?

The integral of (x+2)sin(x^2+4x-6) is -1/2 cos(x^2+4x-6)+C

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integral of (x+2)sin(x^2+4x-6)

\int (x + 2) sin (x^{2} + 4 x - 6) d x

- \frac{1}{2} cos (x^{2} + 4 x - 6) + C

Solution steps

\int (x + 2) sin (x^{2} + 4 x - 6) d x

Apply u-substitution

: \int \frac{sin ( u )}{2} d u

= \int \frac{sin ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{2} (- cos (u))

Substitute back

u = x^{2} + 4 x - 6

= \frac{1}{2} (- cos (x^{2} + 4 x - 6))

Simplify

= - \frac{1}{2} cos (x^{2} + 4 x - 6)

Add a constant to the solution

= - \frac{1}{2} cos (x^{2} + 4 x - 6) + C

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

\frac{\partial}{\partial x} (- y lo g_{10} (x) - (1 - y) lo g_{10} (1 - x))

\int cos^{5} (x) d x

x \to 9 lim (cos (\frac{π x}{4}))

y \to \infty lim (\frac{2 y ^{2} - 3 y + 5}{y ^{2} - 5 y + 2})

What is the integral of (x+2)sin(x^2+4x-6) ?
The integral of (x+2)sin(x^2+4x-6) is -1/2 cos(x^2+4x-6)+C