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

The integral of xsin(-2x) is 1/2 xcos(2x)-1/4 sin(2x)+C

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integral of xsin(-2x)

\int x sin (- 2 x) d x

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

Solution steps

\int x sin (- 2 x) d x

Apply Integration By Parts

: - \int - x^{2} cos (2 x) d x - \frac{1}{2} x^{2} sin (2 x)

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

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

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

Simplify

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

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

Add a constant to the solution

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

9 y^{^{''}} + 24 y^{^{'}} + 16 y = 0

(\frac{1}{1 + x ^{2}})^{^{'}}

\int_{- 1}^{5} 3 x^{2} + 6 x d x

y = 4 x^{3} - 4 x, (- 1, 0)

x \to - 2 lim (- 3)

What is the integral of xsin(-2x) ?
The integral of xsin(-2x) is 1/2 xcos(2x)-1/4 sin(2x)+C