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

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

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

\int_{0}^{\frac{π}{2}} sin (t) d t

1

Solution steps

\int_{0}^{\frac{π}{2}} sin (t) d t

Use the common integral:

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

= [- cos (t)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= 1

\frac{d}{d x} (\frac{e ^{x}}{4 x})

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

9 x, 1, 9

\frac{\partial}{\partial x} (sin^{8} (x - 4 y))

x \to 1 lim (\frac{10 x - 9 - 1}{x - 1})

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