Question$\int y^{3} \cos y d y$ (use tabular method

Studdy Solution
Apply the tabular method. Multiply diagonally and alternate signs:
\begin{align*} &+ (y^3)(\sin y) \\ &- (3y^2)(-\cos y) \\ &+ (6y)(-\sin y) \\ &- (6)(\cos y) \\ \end{align*}
So, the integral becomes:
$y^3 \sin y + 3y^2 \cos y - 6y \sin y - 6 \cos y + C$
where $C$ is the constant of integration.
The solution to the integral is:
$\boxed{y^3 \sin y + 3y^2 \cos y - 6y \sin y - 6 \cos y + C}$