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

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

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

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

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

Solution steps

\int (sin (2 x) + cos (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

x^{- \frac{1}{2}}

y = 3 - x, y = 2 x^{2}, [0, 1]

f (x) = 2 x^{2} + 3 x, (- 2, 2)

\frac{\partial}{\partial x} (x^{3} - 12 x + y^{3} - 27 y)

\int 5 sin (x) - 2 cos (x) d x

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