What is the integral of sin^2(2t)cos(2t) ?

The integral of sin^2(2t)cos(2t) is 1/12 (sin(6t)-3sin(2t)cos(4t))+C

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integral of sin^2(2t)cos(2t)

\int sin^{2} (2 t) cos (2 t) d t

\frac{1}{12} (sin (6 t) - 3 sin (2 t) cos (4 t)) + C

Solution steps

\int sin^{2} (2 t) cos (2 t) d t

Rewrite using trig identities

= \int \frac{1}{2} cos (2 t) (1 - cos (4 t)) d t

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (2 t) (1 - cos (4 t)) d t

Apply Integration By Parts

: \frac{1}{2} sin (2 t) (1 - cos (4 t)) - \int 2 sin (4 t) sin (2 t) d t

= \frac{1}{2} (\frac{1}{2} sin (2 t) (1 - cos (4 t)) - \int 2 sin (4 t) sin (2 t) d t)

\int 2 sin (4 t) sin (2 t) d t = - \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t)

= \frac{1}{2} (\frac{1}{2} sin (2 t) (1 - cos (4 t)) - (- \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t)))

Simplify

\frac{1}{2} (\frac{1}{2} sin (2 t) (1 - cos (4 t)) - (- \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t))) : \frac{1}{12} (sin (6 t) - 3 sin (2 t) cos (4 t))

= \frac{1}{12} (sin (6 t) - 3 sin (2 t) cos (4 t))

Add a constant to the solution

= \frac{1}{12} (sin (6 t) - 3 sin (2 t) cos (4 t)) + C

x \to 5 lim (x 41 - x^{2})

\frac{d}{d v} (u^{2} + v^{2})

y^{^{''''}} = y

\frac{d}{d x} (- 64 cos (4 x))

\int 3 x^{\frac{3}{2}} d x

What is the integral of sin^2(2t)cos(2t) ?
The integral of sin^2(2t)cos(2t) is 1/12 (sin(6t)-3sin(2t)cos(4t))+C