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

The integral of (2+cos(2x))^2 is 1/8 (36x+16sin(2x)+sin(4x))+C

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

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

\frac{1}{8} (36 x + 16 sin (2 x) + sin (4 x)) + C

Solution steps

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

Apply u-substitution

: \int (2 + cos (u))^{2} \frac{1}{2} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int (2 + cos (u))^{2} d u

Expand

(2 + cos (u))^{2} : 4 + 4 cos (u) + cos^{2} (u)

= \frac{1}{2} \cdot \int 4 + 4 cos (u) + cos^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 4 d u + \int 4 cos (u) d u + \int cos^{2} (u) d u)

\int 4 d u = 4 u

\int 4 cos (u) d u = 4 sin (u)

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

= \frac{1}{2} (4 u + 4 sin (u) + \frac{1}{2} (u + \frac{1}{2} sin (2 u)))

Substitute back

u = 2 x

= \frac{1}{2} (4 \cdot 2 x + 4 sin (2 x) + \frac{1}{2} (2 x + \frac{1}{2} sin (2 \cdot 2 x)))

Simplify

\frac{1}{2} (4 \cdot 2 x + 4 sin (2 x) + \frac{1}{2} (2 x + \frac{1}{2} sin (2 \cdot 2 x))) : \frac{1}{8} (36 x + 16 sin (2 x) + sin (4 x))

= \frac{1}{8} (36 x + 16 sin (2 x) + sin (4 x))

Add a constant to the solution

= \frac{1}{8} (36 x + 16 sin (2 x) + sin (4 x)) + C

\int_{- 1}^{2} (2 - x^{2}) - (- x) d x

\frac{\partial}{\partial y} (- 0.27 y^{2} + 3.78 y)

y = \frac{f ( x )}{x ^{4}}

\frac{d}{d x} (e^{x^{2}} - cos (x))

\int \frac{4}{x ^{5}} d x

What is the integral of (2+cos(2x))^2 ?
The integral of (2+cos(2x))^2 is 1/8 (36x+16sin(2x)+sin(4x))+C