What is the integral of (x-3)^2sin(x-3) ?

The integral of (x-3)^2sin(x-3) is -x^2cos(x-3)+6xcos(x-3)+2xsin(x-3)-7cos(x-3)-6sin(x-3)+C

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integral of (x-3)^2sin(x-3)

\int (x - 3)^{2} sin (x - 3) d x

- x^{2} cos (x - 3) + 6 x cos (x - 3) + 2 x sin (x - 3) - 7 cos (x - 3) - 6 sin (x - 3) + C

Solution steps

\int (x - 3)^{2} sin (x - 3) d x

Apply u-substitution

: \int u^{2} sin (u) d u

= \int u^{2} sin (u) d u

Apply Integration By Parts

: - u^{2} cos (u) - \int - 2 u cos (u) d u

= - u^{2} cos (u) - \int - 2 u cos (u) d u

\int - 2 u cos (u) d u = - 2 (u sin (u) + cos (u))

= - u^{2} cos (u) - (- 2 (u sin (u) + cos (u)))

Substitute back

u = x - 3

= - (x - 3)^{2} cos (x - 3) - (- 2 ((x - 3) sin (x - 3) + cos (x - 3)))

Simplify

- (x - 3)^{2} cos (x - 3) - (- 2 ((x - 3) sin (x - 3) + cos (x - 3))) : - x^{2} cos (x - 3) + 6 x cos (x - 3) + 2 x sin (x - 3) - 7 cos (x - 3) - 6 sin (x - 3)

= - x^{2} cos (x - 3) + 6 x cos (x - 3) + 2 x sin (x - 3) - 7 cos (x - 3) - 6 sin (x - 3)

Add a constant to the solution

= - x^{2} cos (x - 3) + 6 x cos (x - 3) + 2 x sin (x - 3) - 7 cos (x - 3) - 6 sin (x - 3) + C

\int 8 tan (8 x) d x

y^{^{'}} = \frac{( x ^{2} )}{y}

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

\int_{3}^{0} (- x^{3} - 2 x^{2} + x - 1) d x

y = \frac{x}{x + 3}, (1, 0.25)

What is the integral of (x-3)^2sin(x-3) ?
The integral of (x-3)^2sin(x-3) is -x^2cos(x-3)+6xcos(x-3)+2xsin(x-3)-7cos(x-3)-6sin(x-3)+C