What is the integral of-4xsin(x+2) ?

The integral of-4xsin(x+2) is 2xcos(2x)-sin(2x)+C

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integral of-4xsin(x+2)

\int - 4 x sin (x + 2) d x

2 x cos (2 x) - sin (2 x) + C

Solution steps

\int - 4 x sin (x + 2) d x

Take the constant out:

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

= - 4 \cdot \int x sin (x + 2) d x

Rewrite using trig identities

= - 4 \cdot \int x \cdot 2 cos (x) sin (x) d x

Take the constant out:

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

= - 4 \cdot 2 \cdot \int x cos (x) sin (x) d x

Rewrite using trig identities

= - 4 \cdot 2 \cdot \int \frac{1}{2} x sin (2 x) d x

Take the constant out:

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

= - 4 \cdot 2 \cdot \frac{1}{2} \cdot \int x sin (2 x) d x

Apply Integration By Parts

: - \frac{1}{2} x cos (2 x) - \int - \frac{1}{2} cos (2 x) d x

= - 4 \cdot 2 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - \int - \frac{1}{2} cos (2 x) d x)

\int - \frac{1}{2} cos (2 x) d x = - \frac{1}{4} sin (2 x)

= - 4 \cdot 2 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x)))

Simplify

- 4 \cdot 2 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x))) : 2 x cos (2 x) - sin (2 x)

= 2 x cos (2 x) - sin (2 x)

Add a constant to the solution

= 2 x cos (2 x) - sin (2 x) + C

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What is the integral of-4xsin(x+2) ?
The integral of-4xsin(x+2) is 2xcos(2x)-sin(2x)+C