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

The integral of (sin(2x)+cos(x))^2 is x-1/8 sin(4x)-4/3 cos^3(x)+1/4 sin(2x)+C

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

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

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

Solution steps

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

Expand

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

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

\int sin^{2} (2 x) d x = \frac{1}{2} (x - \frac{1}{4} sin (4 x))

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

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (sin(2x)+cos(x))^2 ?
The integral of (sin(2x)+cos(x))^2 is x-1/8 sin(4x)-4/3 cos^3(x)+1/4 sin(2x)+C