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

The integral of (3sin(x)+2cos(x))^2 is 13/2 x-5/4 sin(2x)+6sin^2(x)+C

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

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

\frac{13}{2} x - \frac{5}{4} sin (2 x) + 6 sin^{2} (x) + C

Solution steps

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

Expand

(3 sin (x) + 2 cos (x))^{2} : 9 sin^{2} (x) + 12 sin (x) cos (x) + 4 cos^{2} (x)

= \int 9 sin^{2} (x) + 12 sin (x) cos (x) + 4 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 9 sin^{2} (x) d x + \int 12 sin (x) cos (x) d x + \int 4 cos^{2} (x) d x

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

\int 12 sin (x) cos (x) d x = 6 sin^{2} (x)

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

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

Simplify

\frac{9}{2} (x - \frac{1}{2} sin (2 x)) + 6 sin^{2} (x) + 2 x + sin (2 x) : \frac{13}{2} x - \frac{5}{4} sin (2 x) + 6 sin^{2} (x)

= \frac{13}{2} x - \frac{5}{4} sin (2 x) + 6 sin^{2} (x)

Add a constant to the solution

= \frac{13}{2} x - \frac{5}{4} sin (2 x) + 6 sin^{2} (x) + C

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What is the integral of (3sin(x)+2cos(x))^2 ?
The integral of (3sin(x)+2cos(x))^2 is 13/2 x-5/4 sin(2x)+6sin^2(x)+C