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

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

Popular >

integral of xysin(x^2y)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = y x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (2 y^{3})

\int \frac{16 x ^{2}}{( x - 24 ) ( x + 8 ) ^{2}} d x

\int_{0}^{1} \frac{2}{1 - x ^{2}} d x

\frac{\partial}{\partial y} (\frac{x + 1}{y - 1})

\int_{0}^{1} \frac{1}{x ( 1 + x )} d x

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