What is the integral of xsin(3t-3x) ?

The integral of xsin(3t-3x) is 1/9 (3xcos(-3x+3t)+sin(-3x+3t))+C

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integral of xsin(3t-3x)

\int x sin (3 t - 3 x) d x

\frac{1}{9} (3 x cos (- 3 x + 3 t) + sin (- 3 x + 3 t)) + C

Solution steps

\int x sin (3 t - 3 x) d x

Apply u-substitution

= \int \frac{sin ( u ) ( u - 3 t )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sin (u) (u - 3 t) d u

Expand

sin (u) (u - 3 t) : u sin (u) - 3 t sin (u)

= \frac{1}{9} \cdot \int u sin (u) - 3 t sin (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{9} (\int u sin (u) d u - \int 3 t sin (u) d u)

\int u sin (u) d u = - u cos (u) + sin (u)

\int 3 t sin (u) d u = - 3 t cos (u)

= \frac{1}{9} (- u cos (u) + sin (u) - (- 3 t cos (u)))

Substitute back

u = 3 t - 3 x

= \frac{1}{9} (- (3 t - 3 x) cos (3 t - 3 x) + sin (3 t - 3 x) - (- 3 t cos (3 t - 3 x)))

Simplify

\frac{1}{9} (- (3 t - 3 x) cos (3 t - 3 x) + sin (3 t - 3 x) - (- 3 t cos (3 t - 3 x))) : \frac{1}{9} (3 x cos (- 3 x + 3 t) + sin (- 3 x + 3 t))

= \frac{1}{9} (3 x cos (- 3 x + 3 t) + sin (- 3 x + 3 t))

Add a constant to the solution

= \frac{1}{9} (3 x cos (- 3 x + 3 t) + sin (- 3 x + 3 t)) + C

What is the integral of xsin(3t-3x) ?
The integral of xsin(3t-3x) is 1/9 (3xcos(-3x+3t)+sin(-3x+3t))+C