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

The integral of x^2sin(3x+2) is -1/6 x^2cos(6x)+1/108 (6xsin(6x)+cos(6x))+C

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

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

- \frac{1}{6} x^{2} cos (6 x) + \frac{1}{108} (6 x sin (6 x) + cos (6 x)) + C

Solution steps

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

Rewrite using trig identities

= \int x^{2} \cdot 2 cos (3 x) sin (3 x) d x

Take the constant out:

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

= 2 \cdot \int x^{2} cos (3 x) sin (3 x) d x

Rewrite using trig identities

= 2 \cdot \int \frac{1}{2} x^{2} sin (6 x) d x

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int x^{2} sin (6 x) d x

Apply Integration By Parts

: - \frac{1}{6} x^{2} cos (6 x) - \int - \frac{1}{3} x cos (6 x) d x

= 2 \cdot \frac{1}{2} (- \frac{1}{6} x^{2} cos (6 x) - \int - \frac{1}{3} x cos (6 x) d x)

\int - \frac{1}{3} x cos (6 x) d x = - \frac{1}{108} (6 x sin (6 x) + cos (6 x))

= 2 \cdot \frac{1}{2} (- \frac{1}{6} x^{2} cos (6 x) - (- \frac{1}{108} (6 x sin (6 x) + cos (6 x))))

Simplify

2 \cdot \frac{1}{2} (- \frac{1}{6} x^{2} cos (6 x) - (- \frac{1}{108} (6 x sin (6 x) + cos (6 x)))) : - \frac{1}{6} x^{2} cos (6 x) + \frac{1}{108} (6 x sin (6 x) + cos (6 x))

= - \frac{1}{6} x^{2} cos (6 x) + \frac{1}{108} (6 x sin (6 x) + cos (6 x))

Add a constant to the solution

= - \frac{1}{6} x^{2} cos (6 x) + \frac{1}{108} (6 x sin (6 x) + cos (6 x)) + C

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What is the integral of x^2sin(3x+2) ?
The integral of x^2sin(3x+2) is -1/6 x^2cos(6x)+1/108 (6xsin(6x)+cos(6x))+C