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

The integral of sin^2(4x)cos^2(x)4x is

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

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

4 (\frac{1}{2} x (\frac{1}{2} x + \frac{1}{4} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (8 x) - \frac{1}{96} sin (6 x) + \frac{1}{160} sin (10 x)) - \frac{1}{57600} (7200 x^{2} - 3600 cos (2 x) + 225 cos (8 x) + 72 cos (10 x) + 200 cos (6 x))) + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: \frac{1}{2} x (\frac{1}{2} x + \frac{1}{4} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (8 x) - \frac{1}{96} sin (6 x) + \frac{1}{160} sin (10 x)) - \int \frac{1}{2} (\frac{1}{2} x + \frac{1}{4} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (8 x) - \frac{1}{96} sin (6 x) + \frac{1}{160} sin (10 x)) d x

= 4 (\frac{1}{2} x (\frac{1}{2} x + \frac{1}{4} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (8 x) - \frac{1}{96} sin (6 x) + \frac{1}{160} sin (10 x)) - \int \frac{1}{2} (\frac{1}{2} x + \frac{1}{4} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (8 x) - \frac{1}{96} sin (6 x) + \frac{1}{160} sin (10 x)) d x)

\int \frac{1}{2} (\frac{1}{2} x + \frac{1}{4} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (8 x) - \frac{1}{96} sin (6 x) + \frac{1}{160} sin (10 x)) d x = \frac{1}{57600} (7200 x^{2} - 3600 cos (2 x) + 225 cos (8 x) + 72 cos (10 x) + 200 cos (6 x))

= 4 (\frac{1}{2} x (\frac{1}{2} x + \frac{1}{4} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (8 x) - \frac{1}{96} sin (6 x) + \frac{1}{160} sin (10 x)) - \frac{1}{57600} (7200 x^{2} - 3600 cos (2 x) + 225 cos (8 x) + 72 cos (10 x) + 200 cos (6 x)))

Add a constant to the solution

= 4 (\frac{1}{2} x (\frac{1}{2} x + \frac{1}{4} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (8 x) - \frac{1}{96} sin (6 x) + \frac{1}{160} sin (10 x)) - \frac{1}{57600} (7200 x^{2} - 3600 cos (2 x) + 225 cos (8 x) + 72 cos (10 x) + 200 cos (6 x))) + C

\frac{d y}{d x} = 5 x e^{y}, y (0) = 0

\int \frac{x ^{3}}{3 - x ^{4}} d x

\frac{d}{d x} (12 x tanh (x))

y = x^{4} - 5 x^{3} + 2, at x = 2

\frac{\partial}{\partial y} (4 - x^{2} - y^{2})

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