What is the integral of sin^4(3x)cos^2(3x) ?

The integral of sin^4(3x)cos^2(3x) is 1/288 (18x-3sin(6x)-4sin^3(3x)cos(3x)+16sin^5(3x)cos(3x))+C

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integral of sin^4(3x)cos^2(3x)

\int sin^{4} (3 x) cos^{2} (3 x) d x

\frac{1}{288} (18 x - 3 sin (6 x) - 4 sin^{3} (3 x) cos (3 x) + 16 sin^{5} (3 x) cos (3 x)) + C

Solution steps

\int sin^{4} (3 x) cos^{2} (3 x) d x

Apply u-substitution

: \int sin^{4} (u) cos^{2} (u) \frac{1}{3} d u

= \int sin^{4} (u) cos^{2} (u) \frac{1}{3} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Expand

sin^{4} (u) (1 - sin^{2} (u)) : sin^{4} (u) - sin^{6} (u)

= \frac{1}{3} \cdot \int sin^{4} (u) - sin^{6} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int sin^{4} (u) d u - \int sin^{6} (u) d u)

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

\int sin^{6} (u) d u = - \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)))

= \frac{1}{3} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)) - (- \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)))))

Substitute back

u = 3 x

= \frac{1}{3} (- \frac{1}{4} sin^{3} (3 x) cos (3 x) + \frac{3}{8} (3 x - \frac{1}{2} sin (2 \cdot 3 x)) - (- \frac{cos ( 3 x ) sin ^{5} ( 3 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (3 x) cos (3 x) + \frac{3}{8} (3 x - \frac{1}{2} sin (2 \cdot 3 x)))))

Simplify

\frac{1}{3} (- \frac{1}{4} sin^{3} (3 x) cos (3 x) + \frac{3}{8} (3 x - \frac{1}{2} sin (2 \cdot 3 x)) - (- \frac{cos ( 3 x ) sin ^{5} ( 3 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (3 x) cos (3 x) + \frac{3}{8} (3 x - \frac{1}{2} sin (2 \cdot 3 x))))) : \frac{1}{288} (18 x - 3 sin (6 x) - 4 sin^{3} (3 x) cos (3 x) + 16 sin^{5} (3 x) cos (3 x))

= \frac{1}{288} (18 x - 3 sin (6 x) - 4 sin^{3} (3 x) cos (3 x) + 16 sin^{5} (3 x) cos (3 x))

Add a constant to the solution

= \frac{1}{288} (18 x - 3 sin (6 x) - 4 sin^{3} (3 x) cos (3 x) + 16 sin^{5} (3 x) cos (3 x)) + C

\frac{\partial}{\partial x} ((8 + 23 x) y^{t a n (x)})

f (t) = ln (t^{2} + 3)

x \to 0 lim (\frac{1 + 7 x - 1 - 5 x}{x})

f (x) = \frac{1}{x ^{6}}, (3, \frac{1}{729})

n = 1 \sum \infty \frac{n !}{3 ^{n}}

What is the integral of sin^4(3x)cos^2(3x) ?
The integral of sin^4(3x)cos^2(3x) is 1/288 (18x-3sin(6x)-4sin^3(3x)cos(3x)+16sin^5(3x)cos(3x))+C