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

The integral of sin^4(6x) is 1/6 (-1/4 sin^3(6x)cos(6x)+3/8 (6x-1/2 sin(12x)))+C

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integral of sin^4(6x)

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

\frac{1}{6} (- \frac{1}{4} sin^{3} (6 x) cos (6 x) + \frac{3}{8} (6 x - \frac{1}{2} sin (12 x))) + C

Solution steps

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

Apply u-substitution

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

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

Apply Integral Reduction

= \frac{1}{6} (- \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}{6} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 6 x

= \frac{1}{6} (- \frac{cos ( 6 x ) sin ^{3} ( 6 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (6 x - \frac{1}{2} sin (2 \cdot 6 x)))

Simplify

\frac{1}{6} (- \frac{cos ( 6 x ) sin ^{3} ( 6 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (6 x - \frac{1}{2} sin (2 \cdot 6 x))) : \frac{1}{6} (- \frac{1}{4} sin^{3} (6 x) cos (6 x) + \frac{3}{8} (6 x - \frac{1}{2} sin (12 x)))

= \frac{1}{6} (- \frac{1}{4} sin^{3} (6 x) cos (6 x) + \frac{3}{8} (6 x - \frac{1}{2} sin (12 x)))

Add a constant to the solution

= \frac{1}{6} (- \frac{1}{4} sin^{3} (6 x) cos (6 x) + \frac{3}{8} (6 x - \frac{1}{2} sin (12 x))) + C

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

2 x^{2} y^{^{''}} + 6 x y^{^{'}} + 2 y = 0, y (1) = 2, y^{^{'}} (1) = - 1

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 t e^{3 t}

x \to 2 - lim (\frac{4}{( x - 2 ) ^{5}})

y^{^{'}} - y = 4 sin (3 t)

What is the integral of sin^4(6x) ?
The integral of sin^4(6x) is 1/6 (-1/4 sin^3(6x)cos(6x)+3/8 (6x-1/2 sin(12x)))+C