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(-y+pi/2)+cos(y)

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

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

- cos (- y + \frac{π}{2}) + cos (y)

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

- cos (- y + \frac{π}{2}) + cos (y)

= - cos (- y + \frac{π}{2}) + cos (y)

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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(-y+pi/2)+cos(y)