What is the integral of (1+cos(4x))^2 ?

The integral of (1+cos(4x))^2 is 1/16 (24x+8sin(4x)+sin(8x))+C

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integral of (1+cos(4x))^2

\int (1 + cos (4 x))^{2} d x

\frac{1}{16} (24 x + 8 sin (4 x) + sin (8 x)) + C

Solution steps

\int (1 + cos (4 x))^{2} d x

Apply u-substitution

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

Take the constant out:

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

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

Expand

(1 + cos (u))^{2} : 1 + 2 cos (u) + cos^{2} (u)

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

\int 1 d u = u

\int 2 cos (u) d u = 2 sin (u)

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

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

Substitute back

u = 4 x

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

Simplify

\frac{1}{4} (4 x + 2 sin (4 x) + \frac{1}{2} (4 x + \frac{1}{2} sin (2 \cdot 4 x))) : \frac{1}{16} (24 x + 8 sin (4 x) + sin (8 x))

= \frac{1}{16} (24 x + 8 sin (4 x) + sin (8 x))

Add a constant to the solution

= \frac{1}{16} (24 x + 8 sin (4 x) + sin (8 x)) + C

What is the integral of (1+cos(4x))^2 ?
The integral of (1+cos(4x))^2 is 1/16 (24x+8sin(4x)+sin(8x))+C