What is the integral from 1 to 2 of ysin(xy) ?

The integral from 1 to 2 of ysin(xy) is -cos(2y)+cos(y)

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integral from 1 to 2 of ysin(xy)

\int_{1}^{2} y sin (x y) d x

- cos (2 y) + cos (y)

Solution steps

\int_{1}^{2} y sin (x y) d x

Take the constant out:

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

= y \cdot \int_{1}^{2} sin (y x) d x

Apply u-substitution

: \int_{y}^{y \cdot 2} \frac{sin ( u )}{y} d u

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

Take the constant out:

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

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

Use the common integral:

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

= y \frac{1}{y} [- cos (u)]_{y}^{y \cdot 2}

Simplify

y \frac{1}{y} [- cos (u)]_{y}^{y \cdot 2} : [- cos (u)]_{y}^{2 y}

= [- cos (u)]_{y}^{2 y}

Compute the boundaries:

- cos (2 y) + cos (y)

= - cos (2 y) + cos (y)

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What is the integral from 1 to 2 of ysin(xy) ?
The integral from 1 to 2 of ysin(xy) is -cos(2y)+cos(y)