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

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

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

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

- 1

Solution steps

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

Apply Integration By Parts

: [x sin (2 x - π) - \int - 2 x cos (2 x) d x]_{0}^{\frac{π}{2}}

= [x sin (2 x - π) - \int - 2 x cos (2 x) d x]_{0}^{\frac{π}{2}}

\int - 2 x cos (2 x) d x = - x sin (2 x) - \frac{1}{2} cos (2 x)

= [x sin (2 x - π) - (- x sin (2 x) - \frac{1}{2} cos (2 x))]_{0}^{\frac{π}{2}}

Simplify

[x sin (2 x - π) - (- x sin (2 x) - \frac{1}{2} cos (2 x))]_{0}^{\frac{π}{2}} : [\frac{1}{2} cos (2 x)]_{0}^{\frac{π}{2}}

= [\frac{1}{2} cos (2 x)]_{0}^{\frac{π}{2}}

Compute the boundaries:

- 1

= - 1

k = 1 \sum \infty \frac{1}{k !}

y = \frac{5}{( 2 x ^{3} )} + 2 cos (x)

\int \frac{ln ( x ^{9} )}{x} d x

\frac{d y}{d x} - e^{4 x + y} = 0

\frac{d}{d x} (4 cos (5 x))

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