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

The integral of xsin^2(4x) is 1/128 (32x^2-8xsin(8x)-cos(8x))+C

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integral of xsin^2(4x)

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

\frac{1}{128} (32 x^{2} - 8 x sin (8 x) - cos (8 x)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} x (1 - cos (8 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 (8 x)) d x

Apply Integration By Parts

: x (x - \frac{1}{8} sin (8 x)) - \int x - \frac{1}{8} sin (8 x) d x

= \frac{1}{2} (x (x - \frac{1}{8} sin (8 x)) - \int x - \frac{1}{8} sin (8 x) d x)

\int x - \frac{1}{8} sin (8 x) d x = \frac{x ^{2}}{2} + \frac{1}{64} cos (8 x)

= \frac{1}{2} (x (x - \frac{1}{8} sin (8 x)) - (\frac{x ^{2}}{2} + \frac{1}{64} cos (8 x)))

Simplify

\frac{1}{2} (x (x - \frac{1}{8} sin (8 x)) - (\frac{x ^{2}}{2} + \frac{1}{64} cos (8 x))) : \frac{1}{128} (32 x^{2} - 8 x sin (8 x) - cos (8 x))

= \frac{1}{128} (32 x^{2} - 8 x sin (8 x) - cos (8 x))

Add a constant to the solution

= \frac{1}{128} (32 x^{2} - 8 x sin (8 x) - cos (8 x)) + C

y = ∣ x - 3 ∣, y = \frac{x}{2}, y = 0

\int 7 x^{2} e^{x} d x

4 x^{5} ln (x)

\frac{d}{d x} (3 - 4 x)

- 4

What is the integral of xsin^2(4x) ?
The integral of xsin^2(4x) is 1/128 (32x^2-8xsin(8x)-cos(8x))+C