What is the integral from 0 to 1 of xsin(xy) ?

The integral from 0 to 1 of xsin(xy) is -cos(x)+1

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integral from 0 to 1 of xsin(xy)

\int_{0}^{1} x sin (x y) d y

- cos (x) + 1

Solution steps

\int_{0}^{1} x sin (x y) d y

Take the constant out:

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

= x \cdot \int_{0}^{1} sin (x y) d y

Apply u-substitution

= x \cdot \int_{0}^{x} \frac{sin ( u )}{x} d u

Take the constant out:

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

= x \frac{1}{x} \cdot \int_{0}^{x} sin (u) d u

Use the common integral:

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

= x \frac{1}{x} [- cos (u)]_{0}^{x}

Simplify

x \frac{1}{x} [- cos (u)]_{0}^{x} : [- cos (u)]_{0}^{x}

= [- cos (u)]_{0}^{x}

Compute the boundaries:

- cos (x) + 1

= - cos (x) + 1

What is the integral from 0 to 1 of xsin(xy) ?
The integral from 0 to 1 of xsin(xy) is -cos(x)+1