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

The integral of (sin(3x)+3cos(3x))^2 is 5x-1/12 sin(6x)-cos^2(3x)+3/4 sin(6x)+C

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

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

5 x - \frac{1}{12} sin (6 x) - cos^{2} (3 x) + \frac{3}{4} sin (6 x) + C

Solution steps

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

Expand

(sin (3 x) + 3 cos (3 x))^{2} : sin^{2} (3 x) + 6 sin (3 x) cos (3 x) + 9 cos^{2} (3 x)

= \int sin^{2} (3 x) + 6 sin (3 x) cos (3 x) + 9 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^{2} (3 x) d x + \int 6 sin (3 x) cos (3 x) d x + \int 9 cos^{2} (3 x) d x

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

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

\int 9 cos^{2} (3 x) d x = \frac{9}{2} (x + \frac{1}{6} sin (6 x))

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

Simplify

\frac{1}{2} (x - \frac{1}{6} sin (6 x)) - cos^{2} (3 x) + \frac{9}{2} (x + \frac{1}{6} sin (6 x)) : 5 x - \frac{1}{12} sin (6 x) - cos^{2} (3 x) + \frac{3}{4} sin (6 x)

= 5 x - \frac{1}{12} sin (6 x) - cos^{2} (3 x) + \frac{3}{4} sin (6 x)

Add a constant to the solution

= 5 x - \frac{1}{12} sin (6 x) - cos^{2} (3 x) + \frac{3}{4} sin (6 x) + C

\frac{\partial}{\partial x} (2 x y^{7} + x^{7} y^{9})

\int 6 tan (6 x) d x

d er i v a t i v e y = \frac{1}{x ^{\frac{1}{2}}}

\frac{d y}{d x} = \frac{1 + 2 y ^{2}}{y sin ( x )}

\int \frac{3 x - 2}{1 - 6 x - 9 x ^{2}} d x

What is the integral of (sin(3x)+3cos(3x))^2 ?
The integral of (sin(3x)+3cos(3x))^2 is 5x-1/12 sin(6x)-cos^2(3x)+3/4 sin(6x)+C