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

The integral of cos^4(4x) is 1/4 (cos^3(4x)sin(4x)+3/8 (4x-1/4 sin(16x)))+C

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integral of cos^4(4x)

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

\frac{1}{4} (cos^{3} (4 x) sin (4 x) + \frac{3}{8} (4 x - \frac{1}{4} sin (16 x))) + C

Solution steps

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

Apply u-substitution

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

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

Simplify

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

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

Substitute back

u = 4 x

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

Simplify

= \frac{1}{4} (cos^{3} (4 x) sin (4 x) + \frac{3}{8} (4 x - \frac{1}{4} sin (16 x)))

Add a constant to the solution

= \frac{1}{4} (cos^{3} (4 x) sin (4 x) + \frac{3}{8} (4 x - \frac{1}{4} sin (16 x))) + C

\frac{\partial}{\partial x} (- 2 e^{- 2 x})

\int a^{2} - z^{2} d z

\frac{\partial}{\partial x} (x e^{x} + 2)

y = (5 x + 2)^{5}

\frac{d}{d x} (\frac{3}{4 x ^{\frac{5}{2}}})

What is the integral of cos^4(4x) ?
The integral of cos^4(4x) is 1/4 (cos^3(4x)sin(4x)+3/8 (4x-1/4 sin(16x)))+C