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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\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