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

The integral of x^3sin^2(4x) is

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

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

\frac{1}{4} x^{4} sin^{2} (4 x) + \frac{1}{8} x^{4} cos (8 x) - \frac{1}{16} x^{3} sin (8 x) - \frac{3}{128} x^{2} cos (8 x) + \frac{3}{512} x sin (8 x) + \frac{3}{4096} cos (8 x) + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{4} x^{4} sin^{2} (4 x) - \int x^{4} sin (8 x) d x

= \frac{1}{4} x^{4} sin^{2} (4 x) - \int x^{4} sin (8 x) d x

\int x^{4} sin (8 x) d x = - \frac{1}{8} x^{4} cos (8 x) + \frac{1}{2} (\frac{1}{8} x^{3} sin (8 x) - \frac{3}{8} (- \frac{1}{8} x^{2} cos (8 x) + \frac{1}{256} (8 x sin (8 x) + cos (8 x))))

= \frac{1}{4} x^{4} sin^{2} (4 x) - (- \frac{1}{8} x^{4} cos (8 x) + \frac{1}{2} (\frac{1}{8} x^{3} sin (8 x) - \frac{3}{8} (- \frac{1}{8} x^{2} cos (8 x) + \frac{1}{256} (8 x sin (8 x) + cos (8 x)))))

Simplify

\frac{1}{4} x^{4} sin^{2} (4 x) - (- \frac{1}{8} x^{4} cos (8 x) + \frac{1}{2} (\frac{1}{8} x^{3} sin (8 x) - \frac{3}{8} (- \frac{1}{8} x^{2} cos (8 x) + \frac{1}{256} (8 x sin (8 x) + cos (8 x))))) : \frac{1}{4} x^{4} sin^{2} (4 x) + \frac{1}{8} x^{4} cos (8 x) - \frac{1}{16} x^{3} sin (8 x) - \frac{3}{128} x^{2} cos (8 x) + \frac{3}{512} x sin (8 x) + \frac{3}{4096} cos (8 x)

= \frac{1}{4} x^{4} sin^{2} (4 x) + \frac{1}{8} x^{4} cos (8 x) - \frac{1}{16} x^{3} sin (8 x) - \frac{3}{128} x^{2} cos (8 x) + \frac{3}{512} x sin (8 x) + \frac{3}{4096} cos (8 x)

Add a constant to the solution

= \frac{1}{4} x^{4} sin^{2} (4 x) + \frac{1}{8} x^{4} cos (8 x) - \frac{1}{16} x^{3} sin (8 x) - \frac{3}{128} x^{2} cos (8 x) + \frac{3}{512} x sin (8 x) + \frac{3}{4096} cos (8 x) + C

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