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

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

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integral from 1 to 2 of sin(xy)

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

\frac{- cos ( 2 y ) + cos ( y )}{y}

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Compute the boundaries:

- cos (2 y) + cos (y)

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

Simplify

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

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