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

The integral of (x^2+3)sin(4x) is -1/4 x^2cos(4x)+1/8 xsin(4x)-23/32 cos(4x)+C

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

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

- \frac{1}{4} x^{2} cos (4 x) + \frac{1}{8} x sin (4 x) - \frac{23}{32} cos (4 x) + C

Solution steps

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

Apply Integration By Parts

: - \frac{1}{4} cos (4 x) (x^{2} + 3) - \int - \frac{1}{2} x cos (4 x) d x

= - \frac{1}{4} cos (4 x) (x^{2} + 3) - \int - \frac{1}{2} x cos (4 x) d x

\int - \frac{1}{2} x cos (4 x) d x = - \frac{1}{32} (4 x sin (4 x) + cos (4 x))

= - \frac{1}{4} cos (4 x) (x^{2} + 3) - (- \frac{1}{32} (4 x sin (4 x) + cos (4 x)))

Simplify

- \frac{1}{4} cos (4 x) (x^{2} + 3) - (- \frac{1}{32} (4 x sin (4 x) + cos (4 x))) : - \frac{1}{4} x^{2} cos (4 x) + \frac{1}{8} x sin (4 x) - \frac{23}{32} cos (4 x)

= - \frac{1}{4} x^{2} cos (4 x) + \frac{1}{8} x sin (4 x) - \frac{23}{32} cos (4 x)

Add a constant to the solution

= - \frac{1}{4} x^{2} cos (4 x) + \frac{1}{8} x sin (4 x) - \frac{23}{32} cos (4 x) + C

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What is the integral of (x^2+3)sin(4x) ?
The integral of (x^2+3)sin(4x) is -1/4 x^2cos(4x)+1/8 xsin(4x)-23/32 cos(4x)+C