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

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

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

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

\frac{1}{2} cos^{2} (3 x) (x - \frac{1}{2} sin (2 x)) + \frac{3}{2} (- \frac{1}{6} x cos (6 x) + \frac{1}{36} sin (6 x) - \frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x))) + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{3}{2} sin (6 x) (x - \frac{1}{2} sin (2 x)) d x = - \frac{3}{2} (- \frac{1}{6} x cos (6 x) + \frac{1}{36} sin (6 x) - \frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x)))

= \frac{1}{2} cos^{2} (3 x) (x - \frac{1}{2} sin (2 x)) - (- \frac{3}{2} (- \frac{1}{6} x cos (6 x) + \frac{1}{36} sin (6 x) - \frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x))))

Simplify

= \frac{1}{2} cos^{2} (3 x) (x - \frac{1}{2} sin (2 x)) + \frac{3}{2} (- \frac{1}{6} x cos (6 x) + \frac{1}{36} sin (6 x) - \frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x)))

Add a constant to the solution

= \frac{1}{2} cos^{2} (3 x) (x - \frac{1}{2} sin (2 x)) + \frac{3}{2} (- \frac{1}{6} x cos (6 x) + \frac{1}{36} sin (6 x) - \frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x))) + C

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