What is the integral of ysin(x+y^2) ?

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

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integral of ysin(x+y^2)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int sin (v) d v

= \frac{1}{2} \cdot \int sin (v) d v

Use the common integral:

\int sin (v) d v = - cos (v)

= \frac{1}{2} (- cos (v))

Substitute back

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

Simplify

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

Add a constant to the solution

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

y^{^{'}} = 3 y t

\int_{0}^{5} f (x) d x

x \to 0 + lim (\frac{sin ( 2 x )}{x})

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

f (x) = - 4 cos (2 x)

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