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

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

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

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

0

Solution steps

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

Apply u-substitution

: \int_{- π}^{π} sin (u) \frac{1}{2} d u

= \int_{- π}^{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_{- π}^{π} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{- π}^{π}

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

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\int_{π}^{x} cos (2 x) d x

\int_{0}^{\frac{π}{4}} x^{2} cos (x) d x

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