What is the integral of (x+1)y^2*sin((xy)/3) ?

The integral of (x+1)y^2*sin((xy)/3) is 3(-ycos((yx)/3)-yxcos((yx)/3)+3sin((yx)/3))+C

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integral of (x+1)y^2*sin((xy)/3)

\int (x + 1) y^{2} \cdot sin (\frac{x y}{3}) d x

3 (- y cos (\frac{y x}{3}) - y x cos (\frac{y x}{3}) + 3 sin (\frac{y x}{3})) + C

Solution steps

\int (x + 1) y^{2} sin (\frac{x y}{3}) d x

Take the constant out:

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

= y^{2} \cdot \int (1 + x) sin (\frac{y x}{3}) d x

Apply u-substitution

: \int \frac{3 sin ( u ) ( y + 3 u )}{y ^{2}} d u

= y^{2} \cdot \int \frac{3 sin ( u ) ( y + 3 u )}{y ^{2}} d u

Take the constant out:

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

= y^{2} \frac{3}{y ^{2}} \cdot \int sin (u) (y + 3 u) d u

Expand

sin (u) (y + 3 u) : y sin (u) + 3 u sin (u)

= y^{2} \frac{3}{y ^{2}} \cdot \int y sin (u) + 3 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

= y^{2} \frac{3}{y ^{2}} (\int y sin (u) d u + \int 3 u sin (u) d u)

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

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

= y^{2} \frac{3}{y ^{2}} (- y cos (u) + 3 (- u cos (u) + sin (u)))

Substitute back

u = \frac{y x}{3}

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

Simplify

y^{2} \frac{3}{y ^{2}} (- y cos (\frac{y x}{3}) + 3 (- \frac{y x}{3} cos (\frac{y x}{3}) + sin (\frac{y x}{3}))) : 3 (- y cos (\frac{y x}{3}) - y x cos (\frac{y x}{3}) + 3 sin (\frac{y x}{3}))

= 3 (- y cos (\frac{y x}{3}) - y x cos (\frac{y x}{3}) + 3 sin (\frac{y x}{3}))

Add a constant to the solution

= 3 (- y cos (\frac{y x}{3}) - y x cos (\frac{y x}{3}) + 3 sin (\frac{y x}{3})) + C

x \to 0 - lim (g (t))

\frac{d}{d x} (ln (x y^{2}))

y^{^{''}} + 4 y = 2 sin (2 x)

\int \frac{1}{4 x ^{2} + 9} d x

\frac{d}{d x} (\frac{1}{6 x ^{2}})

What is the integral of (x+1)y^2*sin((xy)/3) ?
The integral of (x+1)y^2*sin((xy)/3) is 3(-ycos((yx)/3)-yxcos((yx)/3)+3sin((yx)/3))+C