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

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

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integral of sin^4(8x)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Reduction

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

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

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

Substitute back

u = 8 x

= \frac{1}{8} (- \frac{cos ( 8 x ) sin ^{3} ( 8 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (8 x - \frac{1}{2} sin (2 \cdot 8 x)))

Simplify

\frac{1}{8} (- \frac{cos ( 8 x ) sin ^{3} ( 8 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (8 x - \frac{1}{2} sin (2 \cdot 8 x))) : \frac{1}{8} (- \frac{1}{4} sin^{3} (8 x) cos (8 x) + 3 x - \frac{3}{16} sin (16 x))

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

Add a constant to the solution

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

\frac{d}{d x} (n x^{n - 1})

\int - x e^{- 4 x} d x

f (x) = x^{3} - 12 x^{2} + 48 x

y^{^{''}} + 2 y^{^{'}} - 8 y = x e^{- x} + e^{- x}

\frac{\partial}{\partial x} (- u (x, y) cos (k x) sin (k y))

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