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

The integral of y-x/(y^2)sin(x/y) is -cos(x/y)+(y^2)/2+C

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integral of y-x/(y^2)sin(x/y)

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

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

Solution steps

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

\frac{x}{y ^{2}} = \frac{x - y ^{2}}{y ^{2}} + 1

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

Apply u-substitution

: \int - \frac{x ( - u sin ( u ) ( \frac{u ^{2} - x}{x} + 1 ) + x )}{u ^{3}} d u

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

Take the constant out:

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

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

Expand

\frac{- u sin ( u ) ( \frac{u ^{2} - x}{x} + 1 ) + x}{u ^{3}} : - \frac{1}{x} sin (u) + \frac{x}{u ^{3}}

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

\int \frac{1}{x} sin (u) d u = - \frac{1}{x} cos (u)

\int \frac{x}{u ^{3}} d u = - \frac{x}{2 u ^{2}}

= - x (- (- \frac{1}{x} cos (u)) - \frac{x}{2 u ^{2}})

Substitute back

u = \frac{x}{y}

= - x - (- \frac{1}{x} cos (\frac{x}{y})) - \frac{x}{2 ( \frac{x}{y} ) ^{2}}

Simplify

- x - (- \frac{1}{x} cos (\frac{x}{y})) - \frac{x}{2 ( \frac{x}{y} ) ^{2}} : - cos (\frac{x}{y}) + \frac{y ^{2}}{2}

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (1 + x^{4} + y^{3})

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

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{\frac{1}{2}})

\frac{d}{d x} (\frac{( 1 - sec ( x ) )}{tan ( x )})

x + \frac{1}{x ^{2}} + x^{3}

What is the integral of y-x/(y^2)sin(x/y) ?
The integral of y-x/(y^2)sin(x/y) is -cos(x/y)+(y^2)/2+C