What is the integral of cos^4(6x) ?

The integral of cos^4(6x) is 1/6 (cos^3(6x)sin(6x)+3/8 (6x-1/4 sin(24x)))+C

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integral of cos^4(6x)

\int cos^{4} (6 x) d x

\frac{1}{6} (cos^{3} (6 x) sin (6 x) + \frac{3}{8} (6 x - \frac{1}{4} sin (24 x))) + C

Solution steps

\int cos^{4} (6 x) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{6} d u

= \int cos^{4} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int cos^{4} (u) d u

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

= \frac{1}{6} \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

: cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u

= \frac{1}{6} (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= \frac{1}{6} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 6 x

= \frac{1}{6} (cos^{3} (6 x) sin (6 x) - (- \frac{3}{8} (6 x - \frac{1}{4} sin (4 \cdot 6 x))))

Simplify

= \frac{1}{6} (cos^{3} (6 x) sin (6 x) + \frac{3}{8} (6 x - \frac{1}{4} sin (24 x)))

Add a constant to the solution

= \frac{1}{6} (cos^{3} (6 x) sin (6 x) + \frac{3}{8} (6 x - \frac{1}{4} sin (24 x))) + C

\frac{d}{d x} (\frac{1}{e} e^{x})

\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2}

\int x^{2} \cdot sin (x^{3}) d x

n = 1 \sum \infty 15 (\frac{1}{100})^{n}

\frac{d}{d x} (\frac{x ^{2}}{2} + \frac{1}{x})

What is the integral of cos^4(6x) ?
The integral of cos^4(6x) is 1/6 (cos^3(6x)sin(6x)+3/8 (6x-1/4 sin(24x)))+C