What is the integral of (x+2)sin(x^2+4x+8) ?

The integral of (x+2)sin(x^2+4x+8) is -1/2 cos(x^2+4x+8)+C

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integral of (x+2)sin(x^2+4x+8)

\int (x + 2) sin (x^{2} + 4 x + 8) d x

- \frac{1}{2} cos (x^{2} + 4 x + 8) + C

Solution steps

\int (x + 2) sin (x^{2} + 4 x + 8) d x

Apply u-substitution

: \int u sin (u^{2} + 4) d u

= \int u sin (u^{2} + 4) d u

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

\frac{1}{2} (- cos ((x + 2)^{2} + 4)) : - \frac{1}{2} cos (x^{2} + 4 x + 8)

= - \frac{1}{2} cos (x^{2} + 4 x + 8)

Add a constant to the solution

= - \frac{1}{2} cos (x^{2} + 4 x + 8) + C

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What is the integral of (x+2)sin(x^2+4x+8) ?
The integral of (x+2)sin(x^2+4x+8) is -1/2 cos(x^2+4x+8)+C