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

The integral of sin^4(2x)*cos^2(2x) is 1/192 (12x-3sin(4x)-4sin^3(2x)cos(2x)+16sin^5(2x)cos(2x))+C

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

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

\frac{1}{192} (12 x - 3 sin (4 x) - 4 sin^{3} (2 x) cos (2 x) + 16 sin^{5} (2 x) cos (2 x)) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{2} \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}{2} \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}{2} (\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}{2} (- \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 = 2 x

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

Simplify

\frac{1}{2} (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (2 \cdot 2 x)) - (- \frac{cos ( 2 x ) sin ^{5} ( 2 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (2 \cdot 2 x))))) : \frac{1}{192} (12 x - 3 sin (4 x) - 4 sin^{3} (2 x) cos (2 x) + 16 sin^{5} (2 x) cos (2 x))

= \frac{1}{192} (12 x - 3 sin (4 x) - 4 sin^{3} (2 x) cos (2 x) + 16 sin^{5} (2 x) cos (2 x))

Add a constant to the solution

= \frac{1}{192} (12 x - 3 sin (4 x) - 4 sin^{3} (2 x) cos (2 x) + 16 sin^{5} (2 x) cos (2 x)) + C

x \to 8 - lim (\frac{∣ x + 8 ∣}{x + 8})

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

\int sin^{(2)} (x) d x

x \to 16 lim (x + x^{2})

4 + t^{6}

What is the integral of sin^4(2x)*cos^2(2x) ?
The integral of sin^4(2x)*cos^2(2x) is 1/192 (12x-3sin(4x)-4sin^3(2x)cos(2x)+16sin^5(2x)cos(2x))+C