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

The integral of (sin^2(3x)+cos(3x))^2 is 7/8 x+2/9 sin^3(3x)-1/12 sin^3(3x)cos(3x)+1/48 sin(6x)+C

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

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

\frac{7}{8} x + \frac{2}{9} sin^{3} (3 x) - \frac{1}{12} sin^{3} (3 x) cos (3 x) + \frac{1}{48} sin (6 x) + C

Solution steps

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

Expand

(sin^{2} (3 x) + cos (3 x))^{2} : sin^{4} (3 x) + 2 sin^{2} (3 x) cos (3 x) + cos^{2} (3 x)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int sin^{4} (3 x) d x = - \frac{1}{12} sin^{3} (3 x) cos (3 x) + \frac{1}{16} (6 x - sin (6 x))

\int 2 sin^{2} (3 x) cos (3 x) d x = \frac{2}{9} sin^{3} (3 x)

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

= - \frac{1}{12} sin^{3} (3 x) cos (3 x) + \frac{1}{16} (6 x - sin (6 x)) + \frac{2}{9} sin^{3} (3 x) + \frac{1}{2} (x + \frac{1}{6} sin (6 x))

Simplify

- \frac{1}{12} sin^{3} (3 x) cos (3 x) + \frac{1}{16} (6 x - sin (6 x)) + \frac{2}{9} sin^{3} (3 x) + \frac{1}{2} (x + \frac{1}{6} sin (6 x)) : \frac{7}{8} x + \frac{2}{9} sin^{3} (3 x) - \frac{1}{12} sin^{3} (3 x) cos (3 x) + \frac{1}{48} sin (6 x)

= \frac{7}{8} x + \frac{2}{9} sin^{3} (3 x) - \frac{1}{12} sin^{3} (3 x) cos (3 x) + \frac{1}{48} sin (6 x)

Add a constant to the solution

= \frac{7}{8} x + \frac{2}{9} sin^{3} (3 x) - \frac{1}{12} sin^{3} (3 x) cos (3 x) + \frac{1}{48} sin (6 x) + C

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What is the integral of (sin^2(3x)+cos(3x))^2 ?
The integral of (sin^2(3x)+cos(3x))^2 is 7/8 x+2/9 sin^3(3x)-1/12 sin^3(3x)cos(3x)+1/48 sin(6x)+C