What is the integral of xsin^2(2x) ?

The integral of xsin^2(2x) is 1/32 (8x^2-4xsin(4x)-cos(4x))+C

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integral of xsin^2(2x)

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

\frac{1}{32} (8 x^{2} - 4 x sin (4 x) - cos (4 x)) + C

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply Integration By Parts

: x (x - \frac{1}{4} sin (4 x)) - \int x - \frac{1}{4} sin (4 x) d x

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

\int x - \frac{1}{4} sin (4 x) d x = \frac{x ^{2}}{2} + \frac{1}{16} cos (4 x)

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

Simplify

\frac{1}{2} (x (x - \frac{1}{4} sin (4 x)) - (\frac{x ^{2}}{2} + \frac{1}{16} cos (4 x))) : \frac{1}{32} (8 x^{2} - 4 x sin (4 x) - cos (4 x))

= \frac{1}{32} (8 x^{2} - 4 x sin (4 x) - cos (4 x))

Add a constant to the solution

= \frac{1}{32} (8 x^{2} - 4 x sin (4 x) - cos (4 x)) + C

\frac{d}{d x} ((x^{2} + 3 x) e^{2 x})

4 t^{3}

x \to \infty + lim (4 x)

[x + (x + sin^{2} (x))^{3}]^{6}

\int x e^{y x} d x

What is the integral of xsin^2(2x) ?
The integral of xsin^2(2x) is 1/32 (8x^2-4xsin(4x)-cos(4x))+C