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

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

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

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

x - \frac{1}{4} cos (4 x) + C

Solution steps

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

Rewrite using trig identities

= \int 1 + sin (4 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 1 d x + \int sin (4 x) d x

\int 1 d x = x

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

= x - \frac{1}{4} cos (4 x)

Add a constant to the solution

= x - \frac{1}{4} cos (4 x) + C

(ln (y))^{7} \frac{d y}{d x} = x^{7} y, y (1) = e^{2}

\frac{( x ^{3} - 5 x + 186 )}{x ^{2} - x - 42}

sin (x), cos (x), \frac{- π}{30}, \frac{25}{12}

y^{^{''}} + y^{^{'}} + 1.25 y = 0, y (0) = 3, y^{^{'}} (0) = 1

\int 23 sec^{- 2} (x t a) n^{3} x d x

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