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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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