What is the integral of (y+1)x^2sin((xy)/2) ?

The integral of (y+1)x^2sin((xy)/2) is 2(-xcos((xy)/2)-xycos((xy)/2)+2sin((xy)/2))+C

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integral of (y+1)x^2sin((xy)/2)

\int (y + 1) x^{2} sin (\frac{x y}{2}) d y

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

Solution steps

\int (y + 1) x^{2} sin (\frac{x y}{2}) d y

Take the constant out:

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

= x^{2} \cdot \int (1 + y) sin (\frac{x y}{2}) d y

Apply u-substitution

= x^{2} \cdot \int \frac{2 sin ( u ) ( x + 2 u )}{x ^{2}} d u

Take the constant out:

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

= x^{2} \frac{2}{x ^{2}} \cdot \int sin (u) (x + 2 u) d u

Expand

sin (u) (x + 2 u) : x sin (u) + 2 u sin (u)

= x^{2} \frac{2}{x ^{2}} \cdot \int x sin (u) + 2 u 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

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

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

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

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

Substitute back

u = \frac{x y}{2}

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

Simplify

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

= 2 (- x cos (\frac{x y}{2}) - x y cos (\frac{x y}{2}) + 2 sin (\frac{x y}{2}))

Add a constant to the solution

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

What is the integral of (y+1)x^2sin((xy)/2) ?
The integral of (y+1)x^2sin((xy)/2) is 2(-xcos((xy)/2)-xycos((xy)/2)+2sin((xy)/2))+C