What is the integral of [cos(4x)+cos(8x)]^2 ?

The integral of [cos(4x)+cos(8x)]^2 is 1/96 (96x+64sin^3(4x)+6sin(8x)+48cos(8x)sin(4x)+3sin(16x))+C

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integral of [cos(4x)+cos(8x)]^2

\int [cos (4 x) + cos (8 x)]^{2} d x

\frac{1}{96} (96 x + 64 sin^{3} (4 x) + 6 sin (8 x) + 48 cos (8 x) sin (4 x) + 3 sin (16 x)) + C

Solution steps

\int [cos (4 x) + cos (8 x)]^{2} d x

Simplify

(cos (4 x) + cos (8 x))^{2} : (cos (4 x) + cos (8 x)) (cos (4 x) + cos (8 x))

= \int (cos (4 x) + cos (8 x)) (cos (4 x) + cos (8 x)) d x

Simplify

(cos (4 x) + cos (8 x)) (cos (4 x) + cos (8 x)) : (cos (4 x) + cos (8 x))^{2}

= \int (cos (4 x) + cos (8 x))^{2} d x

Expand

(cos (4 x) + cos (8 x))^{2} : cos^{2} (4 x) + 2 cos (4 x) cos (8 x) + cos^{2} (8 x)

= \int cos^{2} (4 x) + 2 cos (4 x) cos (8 x) + cos^{2} (8 x) d x

Apply u-substitution

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

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

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

\int 2 cos (u) cos (2 u) d u = 2 (cos (2 u) sin (u) + \frac{4}{3} sin^{3} (u))

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

= \frac{1}{4} (\frac{1}{2} (u + \frac{1}{2} sin (2 u)) + 2 (cos (2 u) sin (u) + \frac{4}{3} sin^{3} (u)) + \frac{1}{2} (u + \frac{1}{4} sin (4 u)))

Substitute back

u = 4 x

= \frac{1}{4} (\frac{1}{2} (4 x + \frac{1}{2} sin (2 \cdot 4 x)) + 2 (cos (2 \cdot 4 x) sin (4 x) + \frac{4}{3} sin^{3} (4 x)) + \frac{1}{2} (4 x + \frac{1}{4} sin (4 \cdot 4 x)))

Simplify

\frac{1}{4} (\frac{1}{2} (4 x + \frac{1}{2} sin (2 \cdot 4 x)) + 2 (cos (2 \cdot 4 x) sin (4 x) + \frac{4}{3} sin^{3} (4 x)) + \frac{1}{2} (4 x + \frac{1}{4} sin (4 \cdot 4 x))) : \frac{1}{96} (96 x + 64 sin^{3} (4 x) + 6 sin (8 x) + 48 cos (8 x) sin (4 x) + 3 sin (16 x))

= \frac{1}{96} (96 x + 64 sin^{3} (4 x) + 6 sin (8 x) + 48 cos (8 x) sin (4 x) + 3 sin (16 x))

Add a constant to the solution

= \frac{1}{96} (96 x + 64 sin^{3} (4 x) + 6 sin (8 x) + 48 cos (8 x) sin (4 x) + 3 sin (16 x)) + C

\frac{d}{d x} (12 x^{\frac{1}{3}})

\frac{d y}{d x} = \frac{1}{3 x ^{2} y ^{2}}

(4 π r^{2})^{^{'}}

\frac{d}{d x} ((x - 1)^{2} (x - 2) (x - 4))

y = \frac{2 x}{( x ^{2} + 1 )}

What is the integral of [cos(4x)+cos(8x)]^2 ?
The integral of [cos(4x)+cos(8x)]^2 is 1/96 (96x+64sin^3(4x)+6sin(8x)+48cos(8x)sin(4x)+3sin(16x))+C