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

The integral from 0 to pi/2 of cos(2x-pi/2) is 1

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

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

1

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

= \int_{0}^{π} sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{π} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{2} \cdot 2

Simplify

= 1

What is the integral from 0 to pi/2 of cos(2x-pi/2) ?
The integral from 0 to pi/2 of cos(2x-pi/2) is 1