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

The integral of (sin(x)+3cos(x))^2 is 5x+2sin(2x)+3sin^2(x)+C

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

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

5 x + 2 sin (2 x) + 3 sin^{2} (x) + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

= 5 x + 2 sin (2 x) + 3 sin^{2} (x)

Add a constant to the solution

= 5 x + 2 sin (2 x) + 3 sin^{2} (x) + C

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What is the integral of (sin(x)+3cos(x))^2 ?
The integral of (sin(x)+3cos(x))^2 is 5x+2sin(2x)+3sin^2(x)+C