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

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

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integral of sin^4(ax)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

Substitute back

u = a x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (10 x e^{- 2 x})

y = x^{2} - 4, y = 3 x

\frac{d}{d x} (\frac{4 x ^{2} - \frac{1}{2} x}{9 x + 8})

\frac{d}{d x} ((1 + \frac{1}{x})^{2})

\int (- x)^{2 n} d x

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