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

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

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

\int_{\frac{π}{2}}^{0} sin (x) d x

- 1

Solution steps

\int_{\frac{π}{2}}^{0} sin (x) d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Use the common integral:

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

= - [- cos (x)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= - 1

\int_{0}^{\frac{π}{2}} cos (2 u) d u

\int_{1}^{e^{4}} \frac{ln ^{2} ( x ^{4} )}{x} d x

\int_{- \frac{2 π}{3}}^{0} sin (x) d x

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

\int_{0}^{a} e^{- x^{2}} d x

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