What is the integral of sin^4(3x)-cos^4(3x) ?

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

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integral of sin^4(3x)-cos^4(3x)

\int sin^{4} (3 x) - cos^{4} (3 x) d x

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

Solution steps

\int sin^{4} (3 x) - cos^{4} (3 x) d x

Apply Integration By Parts

: (sin^{4} (3 x) - cos^{4} (3 x)) x - \int 12 cos (3 x) sin (3 x) x d x

= (sin^{4} (3 x) - cos^{4} (3 x)) x - \int 12 cos (3 x) sin (3 x) x d x

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

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

Simplify

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

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

Add a constant to the solution

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

\int e^{- s t} \cdot sin (t) d t

\int \frac{1}{x ^{\frac{1}{3}} + 1} d x

\frac{d y}{d x} = x e^{2 y}

\frac{1}{( s - 1 ) \cdot ( s - 2 )}

\frac{d}{d x} (6 arcsin (x^{3}))

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