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

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

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

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

0

Solution steps

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

0

= 0

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

\int_{0}^{4} \frac{1}{1 + u} d u

\int_{3}^{x} \frac{sin ( t )}{t} d t

\int_{- 1}^{1} - 2 x^{2} + 2 d x

\int_{0}^{1} sin (x^{4}) d x

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