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

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

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integral of 1/y sin(x/y)+x

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

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x d x = \frac{x ^{2}}{2}

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

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

Add a constant to the solution

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

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