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

The integral of sin^4(4x) is 1/4 (-1/4 sin^3(4x)cos(4x)+3/8 (4x-1/2 sin(8x)))+C

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integral of sin^4(4x)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Reduction

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

Substitute back

u = 4 x

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

Simplify

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

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

Add a constant to the solution

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

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

x y^{^{'}} - y^{^{'}} - (x - 1) cos (x) + y = 0

\frac{d}{d x} ((1 + 2 sin (x)) cos (x))

n = 1 \sum \infty \frac{n}{( n + 1 ) 3 ^{n}}

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

What is the integral of sin^4(4x) ?
The integral of sin^4(4x) is 1/4 (-1/4 sin^3(4x)cos(4x)+3/8 (4x-1/2 sin(8x)))+C