What is the integral of sin^4(9x) ?

The integral of sin^4(9x) is 1/9 (-1/4 sin^3(9x)cos(9x)+3/8 (9x-1/2 sin(18x)))+C

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integral of sin^4(9x)

\int sin^{4} (9 x) d x

\frac{1}{9} (- \frac{1}{4} sin^{3} (9 x) cos (9 x) + \frac{3}{8} (9 x - \frac{1}{2} sin (18 x))) + C

Solution steps

\int sin^{4} (9 x) d x

Apply u-substitution

: \int sin^{4} (u) \frac{1}{9} d u

= \int sin^{4} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{9} (- \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}{9} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 9 x

= \frac{1}{9} (- \frac{cos ( 9 x ) sin ^{3} ( 9 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (9 x - \frac{1}{2} sin (2 \cdot 9 x)))

Simplify

\frac{1}{9} (- \frac{cos ( 9 x ) sin ^{3} ( 9 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (9 x - \frac{1}{2} sin (2 \cdot 9 x))) : \frac{1}{9} (- \frac{1}{4} sin^{3} (9 x) cos (9 x) + \frac{3}{8} (9 x - \frac{1}{2} sin (18 x)))

= \frac{1}{9} (- \frac{1}{4} sin^{3} (9 x) cos (9 x) + \frac{3}{8} (9 x - \frac{1}{2} sin (18 x)))

Add a constant to the solution

= \frac{1}{9} (- \frac{1}{4} sin^{3} (9 x) cos (9 x) + \frac{3}{8} (9 x - \frac{1}{2} sin (18 x))) + C

\frac{e ^{x}}{7 x ^{2}}

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

y d x + 3 x d y = 0

\int \frac{1}{x x ^{2} - 49} d x

x \to + 0 lim (\frac{e ^{- 1 x} - 1}{x})

What is the integral of sin^4(9x) ?
The integral of sin^4(9x) is 1/9 (-1/4 sin^3(9x)cos(9x)+3/8 (9x-1/2 sin(18x)))+C