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

The integral of+x^3sin(2x) is -1/2 x^3cos(2x)+3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x))+C

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integral of+x^3sin(2x)

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

- \frac{1}{2} x^{3} cos (2 x) + \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{3}{2} x^{2} cos (2 x) d x = - \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= - \frac{1}{2} x^{3} cos (2 x) - (- \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)))

Simplify

= - \frac{1}{2} x^{3} cos (2 x) + \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

Add a constant to the solution

= - \frac{1}{2} x^{3} cos (2 x) + \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) + C

\frac{d y}{d t} = t y^{2} + 2 y^{2}

y^{^{'}} - 3 y = 2 sin (3 x) - cos (3 x)

\int \frac{6 x + 7}{x ^{2} + 4 x + 29} d x

\frac{d}{d x} (\frac{sin ^{2} ( x )}{cos ( x )})

(- x + 2), (- x^{2} - 2 x), - 2, - 1

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