What is the integral of (1+cos(2x))^3 ?

The integral of (1+cos(2x))^3 is 1/24 (60x+9sin(4x)+48sin(2x)-4sin^3(2x))+C

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integral of (1+cos(2x))^3

\int (1 + cos (2 x))^{3} d x

\frac{1}{24} (60 x + 9 sin (4 x) + 48 sin (2 x) - 4 sin^{3} (2 x)) + C

Solution steps

\int (1 + cos (2 x))^{3} d x

Apply u-substitution

: \int (1 + cos (u))^{3} \frac{1}{2} d u

= \int (1 + cos (u))^{3} \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 (1 + cos (u))^{3} d u

Expand

(1 + cos (u))^{3} : 1 + 3 cos (u) + 3 cos^{2} (u) + cos^{3} (u)

= \frac{1}{2} \cdot \int 1 + 3 cos (u) + 3 cos^{2} (u) + cos^{3} (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 1 d u + \int 3 cos (u) d u + \int 3 cos^{2} (u) d u + \int cos^{3} (u) d u)

\int 1 d u = u

\int 3 cos (u) d u = 3 sin (u)

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

\int cos^{3} (u) d u = sin (u) - \frac{sin ^{3} ( u )}{3}

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

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} (2 x + 3 sin (2 x) + \frac{3}{2} (2 x + \frac{1}{2} sin (2 \cdot 2 x)) + sin (2 x) - \frac{sin ^{3} ( 2 x )}{3}) : \frac{1}{24} (60 x + 9 sin (4 x) + 48 sin (2 x) - 4 sin^{3} (2 x))

= \frac{1}{24} (60 x + 9 sin (4 x) + 48 sin (2 x) - 4 sin^{3} (2 x))

Add a constant to the solution

= \frac{1}{24} (60 x + 9 sin (4 x) + 48 sin (2 x) - 4 sin^{3} (2 x)) + C

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What is the integral of (1+cos(2x))^3 ?
The integral of (1+cos(2x))^3 is 1/24 (60x+9sin(4x)+48sin(2x)-4sin^3(2x))+C