What is the integral from 0 to pi/2 of sin(x-y) ?

The integral from 0 to pi/2 of sin(x-y) is cos(x-pi/2)-cos(x)

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integral from 0 to pi/2 of sin(x-y)

\int_{0}^{\frac{π}{2}} sin (x - y) d y

cos (x - \frac{π}{2}) - cos (x)

Solution steps

\int_{0}^{\frac{π}{2}} sin (x - y) d y

Apply u-substitution

: \int_{x}^{x - \frac{π}{2}} - sin (u) d u

= \int_{x}^{x - \frac{π}{2}} - sin (u) d u

Take the constant out:

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

= - \int_{x}^{x - \frac{π}{2}} sin (u) d u

Use the common integral:

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

= - [- cos (u)]_{x}^{x - \frac{π}{2}}

Compute the boundaries:

- cos (x - \frac{π}{2}) + cos (x)

= - (- cos (x - \frac{π}{2}) + cos (x))

Simplify

= cos (x - \frac{π}{2}) - cos (x)

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\int_{3}^{4} \frac{1}{( x + y ) ^{2}} d x

What is the integral from 0 to pi/2 of sin(x-y) ?
The integral from 0 to pi/2 of sin(x-y) is cos(x-pi/2)-cos(x)