What is the integral of cos^4(8x) ?

The integral of cos^4(8x) is 1/256 (32cos^3(8x)sin(8x)+3(32x-sin(32x)))+C

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integral of cos^4(8x)

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

\frac{1}{256} (32 cos^{3} (8 x) sin (8 x) + 3 (32 x - sin (32 x))) + C

Solution steps

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

Apply u-substitution

: \int cos^{4} (u) \frac{1}{8} d u

= \int cos^{4} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int cos^{4} (u) d u

Simplify

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

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

Substitute back

u = 8 x

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

Simplify

\frac{1}{8} (cos^{3} (8 x) sin (8 x) - (- \frac{3}{8} (8 x - \frac{1}{4} sin (4 \cdot 8 x)))) : \frac{1}{256} (32 cos^{3} (8 x) sin (8 x) + 3 (32 x - sin (32 x)))

= \frac{1}{256} (32 cos^{3} (8 x) sin (8 x) + 3 (32 x - sin (32 x)))

Add a constant to the solution

= \frac{1}{256} (32 cos^{3} (8 x) sin (8 x) + 3 (32 x - sin (32 x))) + C

x \to 0 lim (\frac{3 x ^{2} + 5 x}{x})

y^{^{''}} - y^{^{'}} - 2 y = e^{2 x}

\int x^{3} x - 2 d x

\frac{d}{d x} (x^{3} - 5 x^{2} + 4 x - 20 - 5 e^{3 x})

\frac{d}{d t} (tan (cos (2 t)))

What is the integral of cos^4(8x) ?
The integral of cos^4(8x) is 1/256 (32cos^3(8x)sin(8x)+3(32x-sin(32x)))+C